Solve 6^2+17^2=x^2

$$6 ^ { 2 } + 17 ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x=5\sqrt{13}\approx 18.027756377$$

$$x=-5\sqrt{13}\approx -18.027756377$$

Steps by Finding Square Root

Calculate 6 to the power of 2 and get 36.

$$36+17^{2}=x^{2}$$

Calculate 17 to the power of 2 and get 289.

$$36+289=x^{2}$$

Add 36 and 289 to get 325.

$$325=x^{2}$$

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

$$x^{2}=325$$

Take the square root of both sides of the equation.

$$x=5\sqrt{13} x=-5\sqrt{13}$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 2 and get 36.

$$36+17^{2}=x^{2}$$

Calculate 17 to the power of 2 and get 289.

$$36+289=x^{2}$$

Add 36 and 289 to get 325.

$$325=x^{2}$$

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

$$x^{2}=325$$

Subtract 325 from both sides.

$$x^{2}-325=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -325 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-325\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-325\right)}}{2}$$

Multiply -4 times -325.

$$x=\frac{0±\sqrt{1300}}{2}$$

Take the square root of 1300.

$$x=\frac{0±10\sqrt{13}}{2}$$

Now solve the equation x=\frac{0±10\sqrt{13}}{2} when ± is plus.

$$x=5\sqrt{13}$$

Now solve the equation x=\frac{0±10\sqrt{13}}{2} when ± is minus.

$$x=-5\sqrt{13}$$

The equation is now solved.

$$x=5\sqrt{13} x=-5\sqrt{13}$$