Solve 11^2=4^2+x^2

$$11 ^ { 2 } = 4 ^ { 2 } + x ^ { 2 }$$

Solve for x

$$x=\sqrt{105}\approx 10.246950766$$

$$x=-\sqrt{105}\approx -10.246950766$$

Steps by Finding Square Root

Calculate 11 to the power of 2 and get 121.

$$121=4^{2}+x^{2}$$

Calculate 4 to the power of 2 and get 16.

$$121=16+x^{2}$$

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

$$16+x^{2}=121$$

Subtract 16 from both sides.

$$x^{2}=121-16$$

Subtract 16 from 121 to get 105.

$$x^{2}=105$$

Take the square root of both sides of the equation.

$$x=\sqrt{105} x=-\sqrt{105}$$

Steps Using the Quadratic Formula

Calculate 11 to the power of 2 and get 121.

$$121=4^{2}+x^{2}$$

Calculate 4 to the power of 2 and get 16.

$$121=16+x^{2}$$

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

$$16+x^{2}=121$$

Subtract 121 from both sides.

$$16+x^{2}-121=0$$

Subtract 121 from 16 to get -105.

$$-105+x^{2}=0$$

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.

$$x^{2}-105=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-105\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-105\right)}}{2}$$

Multiply -4 times -105.

$$x=\frac{0±\sqrt{420}}{2}$$

Take the square root of 420.

$$x=\frac{0±2\sqrt{105}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{105}}{2} when ± is plus.

$$x=\sqrt{105}$$

Now solve the equation x=\frac{0±2\sqrt{105}}{2} when ± is minus.

$$x=-\sqrt{105}$$

The equation is now solved.

$$x=\sqrt{105} x=-\sqrt{105}$$