Solve 5^2=6^2+x^2

$$5 ^ { 2 } = 6 ^ { 2 } + x ^ { 2 }$$

Solve for x (complex solution)

$$x=-\sqrt{11}i\approx -0-3.31662479i$$

$$x=\sqrt{11}i\approx 3.31662479i$$

Steps by Finding Square Root

Calculate 5 to the power of 2 and get 25.

$$25=6^{2}+x^{2}$$

Calculate 6 to the power of 2 and get 36.

$$25=36+x^{2}$$

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

$$36+x^{2}=25$$

Subtract 36 from both sides.

$$x^{2}=25-36$$

Subtract 36 from 25 to get -11.

$$x^{2}=-11$$

The equation is now solved.

$$x=\sqrt{11}i x=-\sqrt{11}i$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$25=6^{2}+x^{2}$$

Calculate 6 to the power of 2 and get 36.

$$25=36+x^{2}$$

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

$$36+x^{2}=25$$

Subtract 25 from both sides.

$$36+x^{2}-25=0$$

Subtract 25 from 36 to get 11.

$$11+x^{2}=0$$

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.

$$x^{2}+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 11}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 11}}{2}$$

Multiply -4 times 11.

$$x=\frac{0±\sqrt{-44}}{2}$$

Take the square root of -44.

$$x=\frac{0±2\sqrt{11}i}{2}$$

Now solve the equation x=\frac{0±2\sqrt{11}i}{2} when ± is plus.

$$x=\sqrt{11}i$$

Now solve the equation x=\frac{0±2\sqrt{11}i}{2} when ± is minus.

$$x=-\sqrt{11}i$$

The equation is now solved.

$$x=\sqrt{11}i x=-\sqrt{11}i$$