Solve x^2=16^2+17^2

$$x ^ { 2 } = 16 ^ { 2 } + 17 ^ { 2 }$$

Solve for x

$$x=\sqrt{545}\approx 23.34523506$$

$$x=-\sqrt{545}\approx -23.34523506$$

Steps by Finding Square Root

Calculate 16 to the power of 2 and get 256.

$$x^{2}=256+17^{2}$$

Calculate 17 to the power of 2 and get 289.

$$x^{2}=256+289$$

Add 256 and 289 to get 545.

$$x^{2}=545$$

Take the square root of both sides of the equation.

$$x=\sqrt{545} x=-\sqrt{545}$$

Steps Using the Quadratic Formula

Calculate 16 to the power of 2 and get 256.

$$x^{2}=256+17^{2}$$

Calculate 17 to the power of 2 and get 289.

$$x^{2}=256+289$$

Add 256 and 289 to get 545.

$$x^{2}=545$$

Subtract 545 from both sides.

$$x^{2}-545=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -545 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-545\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-545\right)}}{2}$$

Multiply -4 times -545.

$$x=\frac{0±\sqrt{2180}}{2}$$

Take the square root of 2180.

$$x=\frac{0±2\sqrt{545}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{545}}{2} when ± is plus.

$$x=\sqrt{545}$$

Now solve the equation x=\frac{0±2\sqrt{545}}{2} when ± is minus.

$$x=-\sqrt{545}$$

The equation is now solved.

$$x=\sqrt{545} x=-\sqrt{545}$$