Solve x^2+25=57

$${ x }^{ 2 } +25=57$$

Solve for x

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

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

Steps by Finding Square Root

Subtract 25 from both sides.

$$x^{2}=57-25$$

Subtract 25 from 57 to get 32.

$$x^{2}=32$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 57 from both sides.

$$x^{2}+25-57=0$$

Subtract 57 from 25 to get -32.

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

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.

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