## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions to the equation $2y^{2}+7y+3=0$ are $y=-3$ and $y=-\frac{1}{2}$.
1. Factor $2y^{2}+7y+3$. The factors are $(y+3)(2y+1)$ because $y\times2y=2y^{2}$(the a value in a2+bx+c) and $3\times1=3$ the c value in the equation. When the factors are FOILED out, they equal the equation $2y^{2}+7y+3$. 2. Set the factors equal to zero: $y+3=0$ and $2y+1=0$ 3. Solve: $y=-3$ $y=-\frac{1}{2}$