Solve x^2+4=45

$$x ^ { 2 } + 4 = 45$$

Solve for x

$$x=\sqrt{41}\approx 6.403124237$$

$$x=-\sqrt{41}\approx -6.403124237$$

Steps by Finding Square Root

Subtract 4 from both sides.

$$x^{2}=45-4$$

Subtract 4 from 45 to get 41.

$$x^{2}=41$$

Take the square root of both sides of the equation.

$$x=\sqrt{41} x=-\sqrt{41}$$

Steps Using the Quadratic Formula

Subtract 45 from both sides.

$$x^{2}+4-45=0$$

Subtract 45 from 4 to get -41.

$$x^{2}-41=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-41\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-41\right)}}{2}$$

Multiply -4 times -41.

$$x=\frac{0±\sqrt{164}}{2}$$

Take the square root of 164.

$$x=\frac{0±2\sqrt{41}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{41}}{2} when ± is plus.

$$x=\sqrt{41}$$

Now solve the equation x=\frac{0±2\sqrt{41}}{2} when ± is minus.

$$x=-\sqrt{41}$$

The equation is now solved.

$$x=\sqrt{41} x=-\sqrt{41}$$