Solve x^2=20^2+15^2

$${ x }^{ 2 } = { 20 }^{ 2 } + { 15 }^{ 2 }$$

Solve for x

$$x=-25$$

$$x=25$$

Steps Using Difference of Squares

Calculate 20 to the power of 2 and get 400.

$$x^{2}=400+15^{2}$$

Calculate 15 to the power of 2 and get 225.

$$x^{2}=400+225$$

Add 400 and 225 to get 625.

$$x^{2}=625$$

Subtract 625 from both sides.

$$x^{2}-625=0$$

Consider x^{2}-625. Rewrite x^{2}-625 as x^{2}-25^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-25\right)\left(x+25\right)=0$$

To find equation solutions, solve x-25=0 and x+25=0.

$$x=25 x=-25$$

Steps by Finding Square Root

Calculate 20 to the power of 2 and get 400.

$$x^{2}=400+15^{2}$$

Calculate 15 to the power of 2 and get 225.

$$x^{2}=400+225$$

Add 400 and 225 to get 625.

$$x^{2}=625$$

Take the square root of both sides of the equation.

$$x=25 x=-25$$

Steps Using the Quadratic Formula

Calculate 20 to the power of 2 and get 400.

$$x^{2}=400+15^{2}$$

Calculate 15 to the power of 2 and get 225.

$$x^{2}=400+225$$

Add 400 and 225 to get 625.

$$x^{2}=625$$

Subtract 625 from both sides.

$$x^{2}-625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-625\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-625\right)}}{2}$$

Multiply -4 times -625.

$$x=\frac{0±\sqrt{2500}}{2}$$

Take the square root of 2500.

$$x=\frac{0±50}{2}$$

Now solve the equation x=\frac{0±50}{2} when ± is plus. Divide 50 by 2.

$$x=25$$

Now solve the equation x=\frac{0±50}{2} when ± is minus. Divide -50 by 2.

$$x=-25$$

The equation is now solved.

$$x=25 x=-25$$