Solve t^2=7^2+8^2

$$t ^ { 2 } = 7 ^ { 2 } + 8 ^ { 2 }$$

Solve for t

$$t=\sqrt{113}\approx 10.630145813$$

$$t=-\sqrt{113}\approx -10.630145813$$

Steps by Finding Square Root

Calculate 7 to the power of 2 and get 49.

$$t^{2}=49+8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$t^{2}=49+64$$

Add 49 and 64 to get 113.

$$t^{2}=113$$

Take the square root of both sides of the equation.

$$t=\sqrt{113} t=-\sqrt{113}$$

Steps Using the Quadratic Formula

Calculate 7 to the power of 2 and get 49.

$$t^{2}=49+8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$t^{2}=49+64$$

Add 49 and 64 to get 113.

$$t^{2}=113$$

Subtract 113 from both sides.

$$t^{2}-113=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -113 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\left(-113\right)}}{2}$$

Square 0.

$$t=\frac{0±\sqrt{-4\left(-113\right)}}{2}$$

Multiply -4 times -113.

$$t=\frac{0±\sqrt{452}}{2}$$

Take the square root of 452.

$$t=\frac{0±2\sqrt{113}}{2}$$

Now solve the equation t=\frac{0±2\sqrt{113}}{2} when ± is plus.

$$t=\sqrt{113}$$

Now solve the equation t=\frac{0±2\sqrt{113}}{2} when ± is minus.

$$t=-\sqrt{113}$$

The equation is now solved.

$$t=\sqrt{113} t=-\sqrt{113}$$