# How do you solve y^2+3/2y=10?

Jul 6, 2018

color(violet)(y = -4, 5/2

#### Explanation:

${y}^{2} + \left(\frac{3}{2}\right) y = 10$

$2 {y}^{2} + 3 x = 20$

$2 {y}^{2} + 3 x - 20 = 0$

$2 {y}^{2} + 8 y - 5 y - 20 = 0$

$2 y \left(y + 4\right) - 5 \left(y + 4\right) = 0$

$\left(y + 4\right) \cdot \left(2 y - 5\right) = 0$

$y = - 4 , \frac{5}{2}$

Jul 6, 2018

$y = - 4 \text{ or } y = \frac{5}{2}$

#### Explanation:

$\text{subtract 10 from both sides}$

${y}^{2} + \frac{3}{2} y - 10 = 0$

$\text{multiply through by 2}$

$2 {y}^{2} + 3 y - 20 = 0$

$\text{factor the quadratic using the a-c method}$

$\text{the factors of the product } 2 \times - 20 = - 40$

$\text{which sum to "+3" are "+8" and } - 5$

$\text{split the middle term using these factors}$

$2 {y}^{2} + 8 y - 5 y - 20 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$\textcolor{red}{2 y} \left(y + 4\right) \textcolor{red}{- 5} \left(y + 4\right) = 0$

$\text{take out the "color(blue)"common factor } \left(y + 4\right)$

$\left(y + 4\right) \left(\textcolor{red}{2 y - 5}\right) = 0$

$\text{equate each factor to zero and solve for y}$

$y + 4 = 0 \Rightarrow y = - 4$

$2 y - 5 = 0 \Rightarrow y = \frac{5}{2}$