Solve left(3x-8right)^2=3x^2-8x

$${ \left(3x-8 \right) }^{ 2 } =3 { x }^{ 2 } -8x$$

Solve for x

$$x=4$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-8\right)^{2}.

$$9x^{2}-48x+64=3x^{2}-8x$$

Subtract 3x^{2} from both sides.

$$9x^{2}-48x+64-3x^{2}=-8x$$

Combine 9x^{2} and -3x^{2} to get 6x^{2}.

$$6x^{2}-48x+64=-8x$$

Add 8x to both sides.

$$6x^{2}-48x+64+8x=0$$

Combine -48x and 8x to get -40x.

$$6x^{2}-40x+64=0$$

Divide both sides by 2.

$$3x^{2}-20x+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+32. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\times 32=96$$

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 96.

$$-1,-96 -2,-48 -3,-32 -4,-24 -6,-16 -8,-12$$

Calculate the sum for each pair.

$$-1-96=-97 -2-48=-50 -3-32=-35 -4-24=-28 -6-16=-22 -8-12=-20$$

The solution is the pair that gives sum -20.

$$a=-12 b=-8$$

Rewrite 3x^{2}-20x+32 as \left(3x^{2}-12x\right)+\left(-8x+32\right).

$$\left(3x^{2}-12x\right)+\left(-8x+32\right)$$

Factor out 3x in the first and -8 in the second group.

$$3x\left(x-4\right)-8\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(3x-8\right)$$

To find equation solutions, solve x-4=0 and 3x-8=0.

$$x=4 x=\frac{8}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-8\right)^{2}.

$$9x^{2}-48x+64=3x^{2}-8x$$

Subtract 3x^{2} from both sides.

$$9x^{2}-48x+64-3x^{2}=-8x$$

Combine 9x^{2} and -3x^{2} to get 6x^{2}.

$$6x^{2}-48x+64=-8x$$

Add 8x to both sides.

$$6x^{2}-48x+64+8x=0$$

Combine -48x and 8x to get -40x.

$$6x^{2}-40x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -40 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 6\times 64}}{2\times 6}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 6\times 64}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-40\right)±\sqrt{1600-24\times 64}}{2\times 6}$$

Multiply -24 times 64.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1536}}{2\times 6}$$

Add 1600 to -1536.

$$x=\frac{-\left(-40\right)±\sqrt{64}}{2\times 6}$$

Take the square root of 64.

$$x=\frac{-\left(-40\right)±8}{2\times 6}$$

The opposite of -40 is 40.

$$x=\frac{40±8}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{40±8}{12}$$

Now solve the equation x=\frac{40±8}{12} when ± is plus. Add 40 to 8.

$$x=\frac{48}{12}$$

Divide 48 by 12.

$$x=4$$

Now solve the equation x=\frac{40±8}{12} when ± is minus. Subtract 8 from 40.

$$x=\frac{32}{12}$$

Reduce the fraction \frac{32}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{8}{3}$$

The equation is now solved.

$$x=4 x=\frac{8}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-8\right)^{2}.

$$9x^{2}-48x+64=3x^{2}-8x$$

Subtract 3x^{2} from both sides.

$$9x^{2}-48x+64-3x^{2}=-8x$$

Combine 9x^{2} and -3x^{2} to get 6x^{2}.

$$6x^{2}-48x+64=-8x$$

Add 8x to both sides.

$$6x^{2}-48x+64+8x=0$$

Combine -48x and 8x to get -40x.

$$6x^{2}-40x+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}-40x=-64$$

Divide both sides by 6.

$$\frac{6x^{2}-40x}{6}=-\frac{64}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{40}{6}\right)x=-\frac{64}{6}$$

Reduce the fraction \frac{-40}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{20}{3}x=-\frac{64}{6}$$

Reduce the fraction \frac{-64}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{20}{3}x=-\frac{32}{3}$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=-\frac{32}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=-\frac{32}{3}+\frac{100}{9}$$

Add -\frac{32}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{2}{3} x-\frac{10}{3}=-\frac{2}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=4 x=\frac{8}{3}$$