Solve x^2+6^3=498

$$x ^ { 2 } + 6 ^ { 3 } = 498$$

Solve for x

$$x=\sqrt{282}\approx 16.792855624$$

$$x=-\sqrt{282}\approx -16.792855624$$

Steps by Finding Square Root

Calculate 6 to the power of 3 and get 216.

$$x^{2}+216=498$$

Subtract 216 from both sides.

$$x^{2}=498-216$$

Subtract 216 from 498 to get 282.

$$x^{2}=282$$

Take the square root of both sides of the equation.

$$x=\sqrt{282} x=-\sqrt{282}$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 3 and get 216.

$$x^{2}+216=498$$

Subtract 498 from both sides.

$$x^{2}+216-498=0$$

Subtract 498 from 216 to get -282.

$$x^{2}-282=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -282 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-282\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-282\right)}}{2}$$

Multiply -4 times -282.

$$x=\frac{0±\sqrt{1128}}{2}$$

Take the square root of 1128.

$$x=\frac{0±2\sqrt{282}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{282}}{2} when ± is plus.

$$x=\sqrt{282}$$

Now solve the equation x=\frac{0±2\sqrt{282}}{2} when ± is minus.

$$x=-\sqrt{282}$$

The equation is now solved.

$$x=\sqrt{282} x=-\sqrt{282}$$