Solve x^2+20^2=22^2

$${ x }^{ 2 } + { 20 }^{ 2 } = { 22 }^{ 2 }$$

Solve for x

$$x=2\sqrt{21}\approx 9.16515139$$

$$x=-2\sqrt{21}\approx -9.16515139$$

Steps by Finding Square Root

Calculate 20 to the power of 2 and get 400.

$$x^{2}+400=22^{2}$$

Calculate 22 to the power of 2 and get 484.

$$x^{2}+400=484$$

Subtract 400 from both sides.

$$x^{2}=484-400$$

Subtract 400 from 484 to get 84.

$$x^{2}=84$$

Take the square root of both sides of the equation.

$$x=2\sqrt{21} x=-2\sqrt{21}$$

Steps Using the Quadratic Formula

Calculate 20 to the power of 2 and get 400.

$$x^{2}+400=22^{2}$$

Calculate 22 to the power of 2 and get 484.

$$x^{2}+400=484$$

Subtract 484 from both sides.

$$x^{2}+400-484=0$$

Subtract 484 from 400 to get -84.

$$x^{2}-84=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -84 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-84\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-84\right)}}{2}$$

Multiply -4 times -84.

$$x=\frac{0±\sqrt{336}}{2}$$

Take the square root of 336.

$$x=\frac{0±4\sqrt{21}}{2}$$

Now solve the equation x=\frac{0±4\sqrt{21}}{2} when ± is plus.

$$x=2\sqrt{21}$$

Now solve the equation x=\frac{0±4\sqrt{21}}{2} when ± is minus.

$$x=-2\sqrt{21}$$

The equation is now solved.

$$x=2\sqrt{21} x=-2\sqrt{21}$$