Solve 3x^2+7=20

$$3 x ^ { 2 } + 7 = 20$$

Solve for x

$$x = \frac{\sqrt{39}}{3} \approx 2.081665999$$

$$x = -\frac{\sqrt{39}}{3} \approx -2.081665999$$

Steps by Finding Square Root

Subtract 7 from both sides.

$$3x^{2}=20-7$$

Subtract 7 from 20 to get 13.

$$3x^{2}=13$$

Divide both sides by 3.

$$x^{2}=\frac{13}{3}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{39}}{3} x=-\frac{\sqrt{39}}{3}$$

Steps Using the Quadratic Formula

Subtract 20 from both sides.

$$3x^{2}+7-20=0$$

Subtract 20 from 7 to get -13.

$$3x^{2}-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-13\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-13\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-13\right)}}{2\times 3}$$

Multiply -12 times -13.

$$x=\frac{0±\sqrt{156}}{2\times 3}$$

Take the square root of 156.

$$x=\frac{0±2\sqrt{39}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±2\sqrt{39}}{6}$$

Now solve the equation x=\frac{0±2\sqrt{39}}{6} when ± is plus.

$$x=\frac{\sqrt{39}}{3}$$

Now solve the equation x=\frac{0±2\sqrt{39}}{6} when ± is minus.

$$x=-\frac{\sqrt{39}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{39}}{3} x=-\frac{\sqrt{39}}{3}$$