Solve a^2+20^2=25^2

$$a ^ { 2 } + 20 ^ { 2 } = 25 ^ { 2 }$$

Solve for a

$$a=-15$$

$$a=15$$

Steps Using Difference of Squares

Calculate 20 to the power of 2 and get 400.

$$a^{2}+400=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$a^{2}+400=625$$

Subtract 625 from both sides.

$$a^{2}+400-625=0$$

Subtract 625 from 400 to get -225.

$$a^{2}-225=0$$

Consider a^{2}-225. Rewrite a^{2}-225 as a^{2}-15^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(a-15\right)\left(a+15\right)=0$$

To find equation solutions, solve a-15=0 and a+15=0.

$$a=15 a=-15$$

Steps by Finding Square Root

Calculate 20 to the power of 2 and get 400.

$$a^{2}+400=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$a^{2}+400=625$$

Subtract 400 from both sides.

$$a^{2}=625-400$$

Subtract 400 from 625 to get 225.

$$a^{2}=225$$

Take the square root of both sides of the equation.

$$a=15 a=-15$$

Steps Using the Quadratic Formula

Calculate 20 to the power of 2 and get 400.

$$a^{2}+400=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$a^{2}+400=625$$

Subtract 625 from both sides.

$$a^{2}+400-625=0$$

Subtract 625 from 400 to get -225.

$$a^{2}-225=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-225\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-225\right)}}{2}$$

Multiply -4 times -225.

$$a=\frac{0±\sqrt{900}}{2}$$

Take the square root of 900.

$$a=\frac{0±30}{2}$$

Now solve the equation a=\frac{0±30}{2} when ± is plus. Divide 30 by 2.

$$a=15$$

Now solve the equation a=\frac{0±30}{2} when ± is minus. Divide -30 by 2.

$$a=-15$$

The equation is now solved.

$$a=15 a=-15$$