Solve 25+x^2=100

$$25 + x ^ { 2 } = 100$$

Solve for x

$$x=5\sqrt{3}\approx 8.660254038$$

$$x=-5\sqrt{3}\approx -8.660254038$$

Steps by Finding Square Root

Subtract 25 from both sides.

$$x^{2}=100-25$$

Subtract 25 from 100 to get 75.

$$x^{2}=75$$

Take the square root of both sides of the equation.

$$x=5\sqrt{3} x=-5\sqrt{3}$$

Steps Using the Quadratic Formula

Subtract 100 from both sides.

$$25+x^{2}-100=0$$

Subtract 100 from 25 to get -75.

$$-75+x^{2}=0$$

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.

$$x^{2}-75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-75\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-75\right)}}{2}$$

Multiply -4 times -75.

$$x=\frac{0±\sqrt{300}}{2}$$

Take the square root of 300.

$$x=\frac{0±10\sqrt{3}}{2}$$

Now solve the equation x=\frac{0±10\sqrt{3}}{2} when ± is plus.

$$x=5\sqrt{3}$$

Now solve the equation x=\frac{0±10\sqrt{3}}{2} when ± is minus.

$$x=-5\sqrt{3}$$

The equation is now solved.

$$x=5\sqrt{3} x=-5\sqrt{3}$$