Solve (3x-4)^2-8x^2=88+3x^2

$$( 3 x - 4 ) ^ { 2 } - 8 x ^ { 2 } = 88 + 3 x ^ { 2 }$$

Solve for x

$$x=-6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16-8x^{2}=88+3x^{2}$$

Combine 9x^{2} and -8x^{2} to get x^{2}.

$$x^{2}-24x+16=88+3x^{2}$$

Subtract 88 from both sides.

$$x^{2}-24x+16-88=3x^{2}$$

Subtract 88 from 16 to get -72.

$$x^{2}-24x-72=3x^{2}$$

Subtract 3x^{2} from both sides.

$$x^{2}-24x-72-3x^{2}=0$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-24x-72=0$$

Divide both sides by 2.

$$-x^{2}-12x-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-36. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-\left(-36\right)=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -12.

$$a=-6 b=-6$$

Rewrite -x^{2}-12x-36 as \left(-x^{2}-6x\right)+\left(-6x-36\right).

$$\left(-x^{2}-6x\right)+\left(-6x-36\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(-x-6\right)+6\left(-x-6\right)$$

Factor out common term -x-6 by using distributive property.

$$\left(-x-6\right)\left(x+6\right)$$

To find equation solutions, solve -x-6=0 and x+6=0.

$$x=-6 x=-6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16-8x^{2}=88+3x^{2}$$

Combine 9x^{2} and -8x^{2} to get x^{2}.

$$x^{2}-24x+16=88+3x^{2}$$

Subtract 88 from both sides.

$$x^{2}-24x+16-88=3x^{2}$$

Subtract 88 from 16 to get -72.

$$x^{2}-24x-72=3x^{2}$$

Subtract 3x^{2} from both sides.

$$x^{2}-24x-72-3x^{2}=0$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-24x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -24 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-2\right)\left(-72\right)}}{2\left(-2\right)}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\left(-2\right)\left(-72\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-24\right)±\sqrt{576+8\left(-72\right)}}{2\left(-2\right)}$$

Multiply 8 times -72.

$$x=\frac{-\left(-24\right)±\sqrt{576-576}}{2\left(-2\right)}$$

Add 576 to -576.

$$x=\frac{-\left(-24\right)±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$x=-\frac{-24}{2\left(-2\right)}$$

The opposite of -24 is 24.

$$x=\frac{24}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{24}{-4}$$

Divide 24 by -4.

$$x=-6$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16-8x^{2}=88+3x^{2}$$

Combine 9x^{2} and -8x^{2} to get x^{2}.

$$x^{2}-24x+16=88+3x^{2}$$

Subtract 3x^{2} from both sides.

$$x^{2}-24x+16-3x^{2}=88$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-24x+16=88$$

Subtract 16 from both sides.

$$-2x^{2}-24x=88-16$$

Subtract 16 from 88 to get 72.

$$-2x^{2}-24x=72$$

Divide both sides by -2.

$$\frac{-2x^{2}-24x}{-2}=\frac{72}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{24}{-2}\right)x=\frac{72}{-2}$$

Divide -24 by -2.

$$x^{2}+12x=\frac{72}{-2}$$

Divide 72 by -2.

$$x^{2}+12x=-36$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=-36+6^{2}$$

Square 6.

$$x^{2}+12x+36=-36+36$$

Add -36 to 36.

$$x^{2}+12x+36=0$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+6=0 x+6=0$$

Subtract 6 from both sides of the equation.

$$x=-6 x=-6$$

The equation is now solved. Solutions are the same.

$$x=-6$$