Solve 3n^2=7+4

$$3 n ^ { 2 } = 7 + 4$$

Solve for n

$$n = \frac{\sqrt{33}}{3} \approx 1.914854216$$

$$n = -\frac{\sqrt{33}}{3} \approx -1.914854216$$

Steps by Finding Square Root

Add 7 and 4 to get 11.

$$3n^{2}=11$$

Divide both sides by 3.

$$n^{2}=\frac{11}{3}$$

Take the square root of both sides of the equation.

$$n=\frac{\sqrt{33}}{3} n=-\frac{\sqrt{33}}{3}$$

Steps Using the Quadratic Formula

Add 7 and 4 to get 11.

$$3n^{2}=11$$

Subtract 11 from both sides.

$$3n^{2}-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{0±\sqrt{0^{2}-4\times 3\left(-11\right)}}{2\times 3}$$

Square 0.

$$n=\frac{0±\sqrt{-4\times 3\left(-11\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{0±\sqrt{-12\left(-11\right)}}{2\times 3}$$

Multiply -12 times -11.

$$n=\frac{0±\sqrt{132}}{2\times 3}$$

Take the square root of 132.

$$n=\frac{0±2\sqrt{33}}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{0±2\sqrt{33}}{6}$$

Now solve the equation n=\frac{0±2\sqrt{33}}{6} when ± is plus.

$$n=\frac{\sqrt{33}}{3}$$

Now solve the equation n=\frac{0±2\sqrt{33}}{6} when ± is minus.

$$n=-\frac{\sqrt{33}}{3}$$

The equation is now solved.

$$n=\frac{\sqrt{33}}{3} n=-\frac{\sqrt{33}}{3}$$