Solve 2x(x-5)=(x-2)^2

$$2 x ( x - 5 ) = ( x - 2 ) ^ { 2 }$$

Solve for x

$$x=\sqrt{13}+3\approx 6.605551275$$

$$x=3-\sqrt{13}\approx -0.605551275$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-5.

$$2x^{2}-10x=\left(x-2\right)^{2}$$

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

$$2x^{2}-10x=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$2x^{2}-10x-x^{2}=-4x+4$$

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

$$x^{2}-10x=-4x+4$$

Add 4x to both sides.

$$x^{2}-10x+4x=4$$

Combine -10x and 4x to get -6x.

$$x^{2}-6x=4$$

Subtract 4 from both sides.

$$x^{2}-6x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-\left(-6\right)±\sqrt{36+16}}{2}$$

Add 36 to 16.

$$x=\frac{-\left(-6\right)±\sqrt{52}}{2}$$

Take the square root of 52.

$$x=\frac{-\left(-6\right)±2\sqrt{13}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{13}}{2}$$

Now solve the equation x=\frac{6±2\sqrt{13}}{2} when ± is plus. Add 6 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+6}{2}$$

Divide 6+2\sqrt{13} by 2.

$$x=\sqrt{13}+3$$

Now solve the equation x=\frac{6±2\sqrt{13}}{2} when ± is minus. Subtract 2\sqrt{13} from 6.

$$x=\frac{6-2\sqrt{13}}{2}$$

Divide 6-2\sqrt{13} by 2.

$$x=3-\sqrt{13}$$

The equation is now solved.

$$x=\sqrt{13}+3 x=3-\sqrt{13}$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-5.

$$2x^{2}-10x=\left(x-2\right)^{2}$$

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

$$2x^{2}-10x=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$2x^{2}-10x-x^{2}=-4x+4$$

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

$$x^{2}-10x=-4x+4$$

Add 4x to both sides.

$$x^{2}-10x+4x=4$$

Combine -10x and 4x to get -6x.

$$x^{2}-6x=4$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=4+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=4+9$$

Add 4 to 9.

$$x^{2}-6x+9=13$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{13}$$

Simplify.

$$x-3=\sqrt{13} x-3=-\sqrt{13}$$

Add 3 to both sides of the equation.

$$x=\sqrt{13}+3 x=3-\sqrt{13}$$