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

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

Solve for x

$$x=4\sqrt{2}\approx 5.656854249$$

$$x=-4\sqrt{2}\approx -5.656854249$$

Steps by Finding Square Root

Calculate 4 to the power of 2 and get 16.

$$16+4^{2}=x^{2}$$

Calculate 4 to the power of 2 and get 16.

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

Add 16 and 16 to get 32.

$$32=x^{2}$$

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

$$x^{2}=32$$

Take the square root of both sides of the equation.

$$x=4\sqrt{2} x=-4\sqrt{2}$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$16+4^{2}=x^{2}$$

Calculate 4 to the power of 2 and get 16.

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

Add 16 and 16 to get 32.

$$32=x^{2}$$

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

$$x^{2}=32$$

Subtract 32 from both sides.

$$x^{2}-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-32\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-32\right)}}{2}$$

Multiply -4 times -32.

$$x=\frac{0±\sqrt{128}}{2}$$

Take the square root of 128.

$$x=\frac{0±8\sqrt{2}}{2}$$

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

$$x=4\sqrt{2}$$

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

$$x=-4\sqrt{2}$$

The equation is now solved.

$$x=4\sqrt{2} x=-4\sqrt{2}$$