Solve -2n^2=6-8n

$$- 2 n ^ { 2 } = 6 - 8 n$$

Solve for n

$$n=1$$

$$n=3$$

Steps Using Factoring By Grouping

Subtract 6 from both sides.

$$-2n^{2}-6=-8n$$

Add 8n to both sides.

$$-2n^{2}-6+8n=0$$

Divide both sides by 2.

$$-n^{2}-3+4n=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-n^{2}+4n-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-3. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

Rewrite -n^{2}+4n-3 as \left(-n^{2}+3n\right)+\left(n-3\right).

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

Factor out -n in -n^{2}+3n.

$$-n\left(n-3\right)+n-3$$

Factor out common term n-3 by using distributive property.

$$\left(n-3\right)\left(-n+1\right)$$

To find equation solutions, solve n-3=0 and -n+1=0.

$$n=3 n=1$$

Steps Using the Quadratic Formula

Subtract 6 from both sides.

$$-2n^{2}-6=-8n$$

Add 8n to both sides.

$$-2n^{2}-6+8n=0$$

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.

$$-2n^{2}+8n-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 8 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-8±\sqrt{8^{2}-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Square 8.

$$n=\frac{-8±\sqrt{64-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$n=\frac{-8±\sqrt{64+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply 8 times -6.

$$n=\frac{-8±\sqrt{64-48}}{2\left(-2\right)}$$

Add 64 to -48.

$$n=\frac{-8±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$n=\frac{-8±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$n=\frac{-8±4}{-4}$$

Now solve the equation n=\frac{-8±4}{-4} when ± is plus. Add -8 to 4.

$$n=-\frac{4}{-4}$$

Divide -4 by -4.

$$n=1$$

Now solve the equation n=\frac{-8±4}{-4} when ± is minus. Subtract 4 from -8.

$$n=-\frac{12}{-4}$$

Divide -12 by -4.

$$n=3$$

The equation is now solved.

$$n=1 n=3$$

Steps for Completing the Square

Add 8n to both sides.

$$-2n^{2}+8n=6$$

Divide both sides by -2.

$$\frac{-2n^{2}+8n}{-2}=\frac{6}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$n^{2}+\frac{8}{-2}n=\frac{6}{-2}$$

Divide 8 by -2.

$$n^{2}-4n=\frac{6}{-2}$$

Divide 6 by -2.

$$n^{2}-4n=-3$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add -3 to 4.

$$n^{2}-4n+4=1$$

Factor n^{2}-4n+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}.

$$\left(n-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$n-2=1 n-2=-1$$

Add 2 to both sides of the equation.

$$n=3 n=1$$