Solve (2x-3)^2=8(2x-3)+65

$$( 2 x - 3 ) ^ { 2 } = 8 ( 2 x - 3 ) + 65$$

Solve for x

$$x=-1$$

$$x=8$$

Steps Using Factoring By Grouping

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

$$4x^{2}-12x+9=8\left(2x-3\right)+65$$

Use the distributive property to multiply 8 by 2x-3.

$$4x^{2}-12x+9=16x-24+65$$

Add -24 and 65 to get 41.

$$4x^{2}-12x+9=16x+41$$

Subtract 16x from both sides.

$$4x^{2}-12x+9-16x=41$$

Combine -12x and -16x to get -28x.

$$4x^{2}-28x+9=41$$

Subtract 41 from both sides.

$$4x^{2}-28x+9-41=0$$

Subtract 41 from 9 to get -32.

$$4x^{2}-28x-32=0$$

Divide both sides by 4.

$$x^{2}-7x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\left(-8\right)=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=-8 b=1$$

Rewrite x^{2}-7x-8 as \left(x^{2}-8x\right)+\left(x-8\right).

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

Factor out x in x^{2}-8x.

$$x\left(x-8\right)+x-8$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+1\right)$$

To find equation solutions, solve x-8=0 and x+1=0.

$$x=8 x=-1$$

Steps Using the Quadratic Formula

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

$$4x^{2}-12x+9=8\left(2x-3\right)+65$$

Use the distributive property to multiply 8 by 2x-3.

$$4x^{2}-12x+9=16x-24+65$$

Add -24 and 65 to get 41.

$$4x^{2}-12x+9=16x+41$$

Subtract 16x from both sides.

$$4x^{2}-12x+9-16x=41$$

Combine -12x and -16x to get -28x.

$$4x^{2}-28x+9=41$$

Subtract 41 from both sides.

$$4x^{2}-28x+9-41=0$$

Subtract 41 from 9 to get -32.

$$4x^{2}-28x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -28 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 4\left(-32\right)}}{2\times 4}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 4\left(-32\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-28\right)±\sqrt{784-16\left(-32\right)}}{2\times 4}$$

Multiply -16 times -32.

$$x=\frac{-\left(-28\right)±\sqrt{784+512}}{2\times 4}$$

Add 784 to 512.

$$x=\frac{-\left(-28\right)±\sqrt{1296}}{2\times 4}$$

Take the square root of 1296.

$$x=\frac{-\left(-28\right)±36}{2\times 4}$$

The opposite of -28 is 28.

$$x=\frac{28±36}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{28±36}{8}$$

Now solve the equation x=\frac{28±36}{8} when ± is plus. Add 28 to 36.

$$x=\frac{64}{8}$$

Divide 64 by 8.

$$x=8$$

Now solve the equation x=\frac{28±36}{8} when ± is minus. Subtract 36 from 28.

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

Divide -8 by 8.

$$x=-1$$

The equation is now solved.

$$x=8 x=-1$$

Steps for Completing the Square

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

$$4x^{2}-12x+9=8\left(2x-3\right)+65$$

Use the distributive property to multiply 8 by 2x-3.

$$4x^{2}-12x+9=16x-24+65$$

Add -24 and 65 to get 41.

$$4x^{2}-12x+9=16x+41$$

Subtract 16x from both sides.

$$4x^{2}-12x+9-16x=41$$

Combine -12x and -16x to get -28x.

$$4x^{2}-28x+9=41$$

Subtract 9 from both sides.

$$4x^{2}-28x=41-9$$

Subtract 9 from 41 to get 32.

$$4x^{2}-28x=32$$

Divide both sides by 4.

$$\frac{4x^{2}-28x}{4}=\frac{32}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{28}{4}\right)x=\frac{32}{4}$$

Divide -28 by 4.

$$x^{2}-7x=\frac{32}{4}$$

Divide 32 by 4.

$$x^{2}-7x=8$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=8+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=8+\frac{49}{4}$$

Add 8 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{81}{4}$$

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{9}{2} x-\frac{7}{2}=-\frac{9}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=8 x=-1$$