6x 2 6x 12 Factored

$half-dozen \exponential{10}{2} + half dozen x - 12 $

half-dozen\left(x-one\correct)\left(10+ii\right)

6\left(10-1\correct)\left(x+ii\right)

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half dozen\left(x^{2}+x-2\correct)

Gene out 6.

a+b=1 ab=i\left(-2\right)=-2

Consider x^{ii}+x-2. Cistron the expression by grouping. First, the expression needs to be rewritten every bit ten^{2}+ax+bx-2. To find a and b, set upward a system to exist solved.

a=-1 b=ii

Since ab is negative, a and b have the reverse signs. Since a+b is positive, the positive number has greater accented value than the negative. The merely such pair is the organisation solution.

\left(x^{ii}-x\right)+\left(2x-2\correct)

Rewrite x^{2}+x-2 as \left(x^{2}-x\right)+\left(2x-ii\right).

x\left(ten-1\right)+2\left(ten-1\correct)

Factor out ten in the offset and 2 in the second grouping.

\left(ten-one\right)\left(x+2\correct)

Factor out common term x-i by using distributive property.

6\left(x-one\right)\left(x+2\correct)

Rewrite the complete factored expression.

6x^{ii}+6x-12=0

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{ane}\correct)\left(x-x_{2}\correct), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

ten=\frac{-6±\sqrt{six^{ii}-4\times 6\left(-12\right)}}{two\times 6}

All equations of the form ax^{ii}+bx+c=0 tin exist solved using the quadratic formula: \frac{-b±\sqrt{b^{ii}-4ac}}{2a}. The quadratic formula gives 2 solutions, ane when ± is addition and one when it is subtraction.

x=\frac{-6±\sqrt{36-4\times vi\left(-12\correct)}}{ii\times half dozen}

Foursquare 6.

x=\frac{-6±\sqrt{36-24\left(-12\right)}}{two\times half dozen}

Multiply -four times 6.

x=\frac{-6±\sqrt{36+288}}{ii\times 6}

Multiply -24 times -12.

10=\frac{-six±\sqrt{324}}{2\times 6}

Add 36 to 288.

10=\frac{-6±18}{two\times half dozen}

Accept the square root of 324.

x=\frac{-6±18}{12}

Multiply 2 times half dozen.

10=\frac{12}{12}

Now solve the equation x=\frac{-six±18}{12} when ± is plus. Add -six to xviii.

x=\frac{-24}{12}

Now solve the equation x=\frac{-half dozen±18}{12} when ± is minus. Subtract 18 from -half-dozen.

6x^{2}+6x-12=6\left(x-1\right)\left(x-\left(-two\correct)\correct)

Cistron the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{ii}\right). Substitute 1 for x_{ane} and -ii for x_{two}.

6x^{2}+6x-12=6\left(x-1\correct)\left(x+2\right)

Simplify all the expressions of the form p-\left(-q\right) to p+q.

x ^ ii +1x -2 = 0

Quadratic equations such as this one tin be solved by a new direct factoring method that does not require gauge work. To use the direct factoring method, the equation must exist in the form x^ii+Bx+C=0.This is accomplished past dividing both sides of the equation by half-dozen

r + s = -1 rs = -2

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(10−r)(x−s) where sum of factors (r+south)=−B and the product of factors rs = C

r = -\frac{one}{2} - u due south = -\frac{one}{2} + u

Two numbers r and southward sum upwardly to -1 exactly when the average of the two numbers is \frac{1}{2}*-i = -\frac{ane}{2}. You tin can also see that the midpoint of r and s corresponds to the centrality of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div fashion='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -ii

To solve for unknown quantity u, substitute these in the production equation rs = -2

\frac{one}{4} - u^2 = -two

Simplify by expanding (a -b) (a + b) = a^ii – b^2

-u^2 = -2-\frac{one}{4} = -\frac{9}{4}

Simplify the expression by subtracting \frac{one}{4} on both sides

u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{two}

Simplify the expression past multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-\frac{1}{2} - \frac{3}{2} = -ii s = -\frac{ane}{2} + \frac{3}{2} = ane

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and south.

6x 2 6x 12 Factored,

Source: https://mathsolver.microsoft.com/en/solve-problem/6%20x%20%5E%20%7B%202%20%7D%20%2B%206%20x%20-%2012

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