Solve 15^2=5^2+x^2

$${ 15 }^{ 2 } = { 5 }^{ 2 } + { x }^{ 2 }$$

Solve for x

$$x=10\sqrt{2}\approx 14.142135624$$

$$x=-10\sqrt{2}\approx -14.142135624$$

Steps by Finding Square Root

Calculate 15 to the power of 2 and get 225.

$$225=5^{2}+x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$225=25+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$25+x^{2}=225$$

Subtract 25 from both sides.

$$x^{2}=225-25$$

Subtract 25 from 225 to get 200.

$$x^{2}=200$$

Take the square root of both sides of the equation.

$$x=10\sqrt{2} x=-10\sqrt{2}$$

Steps Using the Quadratic Formula

Calculate 15 to the power of 2 and get 225.

$$225=5^{2}+x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$225=25+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$25+x^{2}=225$$

Subtract 225 from both sides.

$$25+x^{2}-225=0$$

Subtract 225 from 25 to get -200.

$$-200+x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$x^{2}-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-200\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-200\right)}}{2}$$

Multiply -4 times -200.

$$x=\frac{0±\sqrt{800}}{2}$$

Take the square root of 800.

$$x=\frac{0±20\sqrt{2}}{2}$$

Now solve the equation x=\frac{0±20\sqrt{2}}{2} when ± is plus.

$$x=10\sqrt{2}$$

Now solve the equation x=\frac{0±20\sqrt{2}}{2} when ± is minus.

$$x=-10\sqrt{2}$$

The equation is now solved.

$$x=10\sqrt{2} x=-10\sqrt{2}$$