Solve 5y+2y^2=12

$$5 y + 2 y ^ { 2 } = 12$$

Solve for y

$$y=-4$$

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

Steps Using Factoring By Grouping

Subtract 12 from both sides.

$$5y+2y^{2}-12=0$$

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

$$2y^{2}+5y-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2y^{2}+ay+by-12. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-12\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite 2y^{2}+5y-12 as \left(2y^{2}-3y\right)+\left(8y-12\right).

$$\left(2y^{2}-3y\right)+\left(8y-12\right)$$

Factor out y in the first and 4 in the second group.

$$y\left(2y-3\right)+4\left(2y-3\right)$$

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

$$\left(2y-3\right)\left(y+4\right)$$

To find equation solutions, solve 2y-3=0 and y+4=0.

$$y=\frac{3}{2} y=-4$$

Steps Using the Quadratic Formula

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.

$$2y^{2}+5y=12$$

Subtract 12 from both sides of the equation.

$$2y^{2}+5y-12=12-12$$

Subtracting 12 from itself leaves 0.

$$2y^{2}+5y-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-5±\sqrt{5^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-5±\sqrt{25-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$y=\frac{-5±\sqrt{25+96}}{2\times 2}$$

Add 25 to 96.

$$y=\frac{-5±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$y=\frac{-5±11}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-5±11}{4}$$

Now solve the equation y=\frac{-5±11}{4} when ± is plus. Add -5 to 11.

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

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

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

Now solve the equation y=\frac{-5±11}{4} when ± is minus. Subtract 11 from -5.

$$y=-\frac{16}{4}$$

Divide -16 by 4.

$$y=-4$$

The equation is now solved.

$$y=\frac{3}{2} y=-4$$

Steps for Completing the Square

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.

$$2y^{2}+5y=12$$

Divide both sides by 2.

$$\frac{2y^{2}+5y}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\frac{5}{2}y=\frac{12}{2}$$

Divide 12 by 2.

$$y^{2}+\frac{5}{2}y=6$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{5}{2}y+\left(\frac{5}{4}\right)^{2}=6+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{5}{2}y+\frac{25}{16}=6+\frac{25}{16}$$

Add 6 to \frac{25}{16}.

$$y^{2}+\frac{5}{2}y+\frac{25}{16}=\frac{121}{16}$$

Factor y^{2}+\frac{5}{2}y+\frac{25}{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}.

$$\left(y+\frac{5}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{5}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$y+\frac{5}{4}=\frac{11}{4} y+\frac{5}{4}=-\frac{11}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$y=\frac{3}{2} y=-4$$