Solve (x-6)^2=2x^2-15x+38

$$( x - 6 ) ^ { 2 } = 2 x ^ { 2 } - 15 x + 38$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36=2x^{2}-15x+38$$

Subtract 2x^{2} from both sides.

$$x^{2}-12x+36-2x^{2}=-15x+38$$

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

$$-x^{2}-12x+36=-15x+38$$

Add 15x to both sides.

$$-x^{2}-12x+36+15x=38$$

Combine -12x and 15x to get 3x.

$$-x^{2}+3x+36=38$$

Subtract 38 from both sides.

$$-x^{2}+3x+36-38=0$$

Subtract 38 from 36 to get -2.

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

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

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

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

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

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

$$-x\left(x-2\right)+x-2$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(-x+1\right)$$

To find equation solutions, solve x-2=0 and -x+1=0.

$$x=2 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36=2x^{2}-15x+38$$

Subtract 2x^{2} from both sides.

$$x^{2}-12x+36-2x^{2}=-15x+38$$

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

$$-x^{2}-12x+36=-15x+38$$

Add 15x to both sides.

$$-x^{2}-12x+36+15x=38$$

Combine -12x and 15x to get 3x.

$$-x^{2}+3x+36=38$$

Subtract 38 from both sides.

$$-x^{2}+3x+36-38=0$$

Subtract 38 from 36 to get -2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±1}{-2}$$

Now solve the equation x=\frac{-3±1}{-2} when ± is plus. Add -3 to 1.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-3±1}{-2} when ± is minus. Subtract 1 from -3.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36=2x^{2}-15x+38$$

Subtract 2x^{2} from both sides.

$$x^{2}-12x+36-2x^{2}=-15x+38$$

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

$$-x^{2}-12x+36=-15x+38$$

Add 15x to both sides.

$$-x^{2}-12x+36+15x=38$$

Combine -12x and 15x to get 3x.

$$-x^{2}+3x+36=38$$

Subtract 36 from both sides.

$$-x^{2}+3x=38-36$$

Subtract 36 from 38 to get 2.

$$-x^{2}+3x=2$$

Divide both sides by -1.

$$\frac{-x^{2}+3x}{-1}=\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{3}{-1}x=\frac{2}{-1}$$

Divide 3 by -1.

$$x^{2}-3x=\frac{2}{-1}$$

Divide 2 by -1.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$