Solve x^2=4^2+12^2

$$x ^ { 2 } = 4 ^ { 2 } + 12 ^ { 2 }$$

Solve for x

$$x=4\sqrt{10}\approx 12.649110641$$

$$x=-4\sqrt{10}\approx -12.649110641$$

Steps by Finding Square Root

Calculate 4 to the power of 2 and get 16.

$$x^{2}=16+12^{2}$$

Calculate 12 to the power of 2 and get 144.

$$x^{2}=16+144$$

Add 16 and 144 to get 160.

$$x^{2}=160$$

Take the square root of both sides of the equation.

$$x=4\sqrt{10} x=-4\sqrt{10}$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$x^{2}=16+12^{2}$$

Calculate 12 to the power of 2 and get 144.

$$x^{2}=16+144$$

Add 16 and 144 to get 160.

$$x^{2}=160$$

Subtract 160 from both sides.

$$x^{2}-160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-160\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-160\right)}}{2}$$

Multiply -4 times -160.

$$x=\frac{0±\sqrt{640}}{2}$$

Take the square root of 640.

$$x=\frac{0±8\sqrt{10}}{2}$$

Now solve the equation x=\frac{0±8\sqrt{10}}{2} when ± is plus.

$$x=4\sqrt{10}$$

Now solve the equation x=\frac{0±8\sqrt{10}}{2} when ± is minus.

$$x=-4\sqrt{10}$$

The equation is now solved.

$$x=4\sqrt{10} x=-4\sqrt{10}$$