# How do you factor and solve 2y^2 +7y + 3 =0?

Mar 28, 2017

Factor is $\left(2 y + 1\right) \left(y + 3\right)$
Solving; $y = - \frac{1}{2}$ and $y = - 3$

#### Explanation:

First, make sure the equation is in the standard form
a${x}^{2}$ + bx + c = 0

$2 {y}^{2} + 7 y + 3 = 0$

If $a \ne 1$, the $a c$ method is applied.

In this case, $a = 2$ so we have to use the $a c$ method.

What is the ac method???

The ac method is simply the value of a times the value of c

$a = 2$

$b = 7$

$c = 3$

$a \cdot c = 6$

Now we have to find the numbers that will give a product of 6 as well as a sum of 7.

$6 \cdot 1 = 6 \left(a c\right)$
$6 + 1 = 7 \left(b\right)$

Bingo! Now that we have these numbers, we rewrite the equation substituting $6$ and $1$ for b.

2${y}^{2}$ + y + 6y + 3 = 0 ...this is the same as the inital equation when simplified.

Now, we factor by grouping and pulling out the gcf .

$\left(2 {y}^{2} + y\right) + \left(6 y + 3\right) = 0$

$y \left(2 y + 1\right) + 3 \left(2 y + 1\right) = 0$

$\left(2 y + 1\right) \left(y + 3\right) = 0$

Solve for y
$2 y + 1 = 0$
$2 y = - 1$

Dividing both sides by 2, we get
$\frac{\cancel{2} y}{\cancel{2}} = \frac{- 1}{2}$
$y = - \frac{1}{2}$

Solve the other also for y
$y + 3 = 0$
$y = - 3$