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

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

Solve for x

$$x=-5$$

$$x=5$$

Steps Using Difference of Squares

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

$$2x^{2}+10x=\left(x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$2x^{2}+10x=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$2x^{2}+10x-x^{2}=10x+25$$

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

$$x^{2}+10x=10x+25$$

Subtract 10x from both sides.

$$x^{2}+10x-10x=25$$

Combine 10x and -10x to get 0.

$$x^{2}=25$$

Subtract 25 from both sides.

$$x^{2}-25=0$$

Consider x^{2}-25. Rewrite x^{2}-25 as x^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-5\right)\left(x+5\right)=0$$

To find equation solutions, solve x-5=0 and x+5=0.

$$x=5 x=-5$$

Steps by Finding Square Root

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

$$2x^{2}+10x=\left(x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$2x^{2}+10x=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$2x^{2}+10x-x^{2}=10x+25$$

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

$$x^{2}+10x=10x+25$$

Subtract 10x from both sides.

$$x^{2}+10x-10x=25$$

Combine 10x and -10x to get 0.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

Steps Using the Quadratic Formula

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

$$2x^{2}+10x=\left(x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$2x^{2}+10x=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$2x^{2}+10x-x^{2}=10x+25$$

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

$$x^{2}+10x=10x+25$$

Subtract 10x from both sides.

$$x^{2}+10x-10x=25$$

Combine 10x and -10x to get 0.

$$x^{2}=25$$

Subtract 25 from both sides.

$$x^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$x=\frac{0±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{0±10}{2}$$

Now solve the equation x=\frac{0±10}{2} when ± is plus. Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{0±10}{2} when ± is minus. Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$