Solve -y=2y^2-6

$$- y = 2 y ^ { 2 } - 6$$

Solve for y

$$y=-2$$

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Subtract 2y^{2} from both sides.

$$-y-2y^{2}=-6$$

Add 6 to both sides.

$$-y-2y^{2}+6=0$$

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

$$-2y^{2}-y+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2\times 6=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=3 b=-4$$

Rewrite -2y^{2}-y+6 as \left(-2y^{2}+3y\right)+\left(-4y+6\right).

$$\left(-2y^{2}+3y\right)+\left(-4y+6\right)$$

Factor out -y in the first and -2 in the second group.

$$-y\left(2y-3\right)-2\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(-y-2\right)$$

To find equation solutions, solve 2y-3=0 and -y-2=0.

$$y=\frac{3}{2} y=-2$$

Steps Using the Quadratic Formula

Subtract 2y^{2} from both sides.

$$-y-2y^{2}=-6$$

Add 6 to both sides.

$$-y-2y^{2}+6=0$$

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.

$$-2y^{2}-y+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -1 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$y=\frac{-\left(-1\right)±\sqrt{1+8\times 6}}{2\left(-2\right)}$$

Multiply 8 times 6.

$$y=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-2\right)}$$

Add 1 to 48.

$$y=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-2\right)}$$

Take the square root of 49.

$$y=\frac{-\left(-1\right)±7}{2\left(-2\right)}$$

The opposite of -1 is 1.

$$y=\frac{1±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{1±7}{-4}$$

Now solve the equation y=\frac{1±7}{-4} when ± is plus. Add 1 to 7.

$$y=\frac{8}{-4}$$

Divide 8 by -4.

$$y=-2$$

Now solve the equation y=\frac{1±7}{-4} when ± is minus. Subtract 7 from 1.

$$y=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

$$y=\frac{3}{2}$$

The equation is now solved.

$$y=-2 y=\frac{3}{2}$$

Steps for Completing the Square

Subtract 2y^{2} from both sides.

$$-y-2y^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2y^{2}-y=-6$$

Divide both sides by -2.

$$\frac{-2y^{2}-y}{-2}=-\frac{6}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$y^{2}+\left(-\frac{1}{-2}\right)y=-\frac{6}{-2}$$

Divide -1 by -2.

$$y^{2}+\frac{1}{2}y=-\frac{6}{-2}$$

Divide -6 by -2.

$$y^{2}+\frac{1}{2}y=3$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{1}{2}y+\left(\frac{1}{4}\right)^{2}=3+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{1}{2}y+\frac{1}{16}=3+\frac{1}{16}$$

Add 3 to \frac{1}{16}.

$$y^{2}+\frac{1}{2}y+\frac{1}{16}=\frac{49}{16}$$

Factor y^{2}+\frac{1}{2}y+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$y+\frac{1}{4}=\frac{7}{4} y+\frac{1}{4}=-\frac{7}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$y=\frac{3}{2} y=-2$$