Solve x^2=60^2+40^2

$${ x }^{ 2 } = { 60 }^{ 2 } + { 40 }^{ 2 }$$

Solve for x

$$x=20\sqrt{13}\approx 72.111025509$$

$$x=-20\sqrt{13}\approx -72.111025509$$

Steps by Finding Square Root

Calculate 60 to the power of 2 and get 3600.

$$x^{2}=3600+40^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$x^{2}=3600+1600$$

Add 3600 and 1600 to get 5200.

$$x^{2}=5200$$

Take the square root of both sides of the equation.

$$x=20\sqrt{13} x=-20\sqrt{13}$$

Steps Using the Quadratic Formula

Calculate 60 to the power of 2 and get 3600.

$$x^{2}=3600+40^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$x^{2}=3600+1600$$

Add 3600 and 1600 to get 5200.

$$x^{2}=5200$$

Subtract 5200 from both sides.

$$x^{2}-5200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -5200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-5200\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-5200\right)}}{2}$$

Multiply -4 times -5200.

$$x=\frac{0±\sqrt{20800}}{2}$$

Take the square root of 20800.

$$x=\frac{0±40\sqrt{13}}{2}$$

Now solve the equation x=\frac{0±40\sqrt{13}}{2} when ± is plus.

$$x=20\sqrt{13}$$

Now solve the equation x=\frac{0±40\sqrt{13}}{2} when ± is minus.

$$x=-20\sqrt{13}$$

The equation is now solved.

$$x=20\sqrt{13} x=-20\sqrt{13}$$