Solve 2n^2-10n-5=-4n

$$2 n ^ { 2 } - 10 n - 5 = - 4 n$$

Solve for n

$$n = \frac{\sqrt{19} + 3}{2} \approx 3.679449472$$

$$n=\frac{3-\sqrt{19}}{2}\approx -0.679449472$$

Steps Using the Quadratic Formula

Add 4n to both sides.

$$2n^{2}-10n-5+4n=0$$

Combine -10n and 4n to get -6n.

$$2n^{2}-6n-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square -6.

$$n=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-\left(-6\right)±\sqrt{36-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$n=\frac{-\left(-6\right)±\sqrt{36+40}}{2\times 2}$$

Add 36 to 40.

$$n=\frac{-\left(-6\right)±\sqrt{76}}{2\times 2}$$

Take the square root of 76.

$$n=\frac{-\left(-6\right)±2\sqrt{19}}{2\times 2}$$

The opposite of -6 is 6.

$$n=\frac{6±2\sqrt{19}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{6±2\sqrt{19}}{4}$$

Now solve the equation n=\frac{6±2\sqrt{19}}{4} when ± is plus. Add 6 to 2\sqrt{19}.

$$n=\frac{2\sqrt{19}+6}{4}$$

Divide 6+2\sqrt{19} by 4.

$$n=\frac{\sqrt{19}+3}{2}$$

Now solve the equation n=\frac{6±2\sqrt{19}}{4} when ± is minus. Subtract 2\sqrt{19} from 6.

$$n=\frac{6-2\sqrt{19}}{4}$$

Divide 6-2\sqrt{19} by 4.

$$n=\frac{3-\sqrt{19}}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{19}+3}{2} n=\frac{3-\sqrt{19}}{2}$$

Steps for Completing the Square

Add 4n to both sides.

$$2n^{2}-10n-5+4n=0$$

Combine -10n and 4n to get -6n.

$$2n^{2}-6n-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$2n^{2}-6n=5$$

Divide both sides by 2.

$$\frac{2n^{2}-6n}{2}=\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\left(-\frac{6}{2}\right)n=\frac{5}{2}$$

Divide -6 by 2.

$$n^{2}-3n=\frac{5}{2}$$

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

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=\frac{5}{2}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-3n+\frac{9}{4}=\frac{5}{2}+\frac{9}{4}$$

Add \frac{5}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-3n+\frac{9}{4}=\frac{19}{4}$$

Factor n^{2}-3n+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{3}{2}\right)^{2}=\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{19}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{\sqrt{19}}{2} n-\frac{3}{2}=-\frac{\sqrt{19}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$n=\frac{\sqrt{19}+3}{2} n=\frac{3-\sqrt{19}}{2}$$