Solve 3n^2=2-7n

$$3 n ^ { 2 } = 2 - 7 n$$

Solve for n

$$n=\frac{\sqrt{73}-7}{6}\approx 0.257333958$$

$$n=\frac{-\sqrt{73}-7}{6}\approx -2.590667291$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

$$3n^{2}-2=-7n$$

Add 7n to both sides.

$$3n^{2}-2+7n=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3n^{2}+7n-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-7±\sqrt{7^{2}-4\times 3\left(-2\right)}}{2\times 3}$$

Square 7.

$$n=\frac{-7±\sqrt{49-4\times 3\left(-2\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-7±\sqrt{49-12\left(-2\right)}}{2\times 3}$$

Multiply -12 times -2.

$$n=\frac{-7±\sqrt{49+24}}{2\times 3}$$

Add 49 to 24.

$$n=\frac{-7±\sqrt{73}}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-7±\sqrt{73}}{6}$$

Now solve the equation n=\frac{-7±\sqrt{73}}{6} when ± is plus. Add -7 to \sqrt{73}.

$$n=\frac{\sqrt{73}-7}{6}$$

Now solve the equation n=\frac{-7±\sqrt{73}}{6} when ± is minus. Subtract \sqrt{73} from -7.

$$n=\frac{-\sqrt{73}-7}{6}$$

The equation is now solved.

$$n=\frac{\sqrt{73}-7}{6} n=\frac{-\sqrt{73}-7}{6}$$

Steps for Completing the Square

Add 7n to both sides.

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

Divide both sides by 3.

$$\frac{3n^{2}+7n}{3}=\frac{2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\frac{7}{3}n=\frac{2}{3}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{7}{3}n+\left(\frac{7}{6}\right)^{2}=\frac{2}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{7}{3}n+\frac{49}{36}=\frac{2}{3}+\frac{49}{36}$$

Add \frac{2}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{7}{3}n+\frac{49}{36}=\frac{73}{36}$$

Factor n^{2}+\frac{7}{3}n+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{7}{6}\right)^{2}=\frac{73}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{7}{6}\right)^{2}}=\sqrt{\frac{73}{36}}$$

Simplify.

$$n+\frac{7}{6}=\frac{\sqrt{73}}{6} n+\frac{7}{6}=-\frac{\sqrt{73}}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$n=\frac{\sqrt{73}-7}{6} n=\frac{-\sqrt{73}-7}{6}$$