# How do you solve 5r+3=2r^2?

Aug 8, 2018

${r}_{1} = - \frac{1}{2}$ and ${r}_{2} = 3$

#### Explanation:

$5 r + 3 = 2 {r}^{2}$

$2 {r}^{2} - 5 r - 3 = 0$

$\left(2 r + 1\right) \cdot \left(r - 3\right) = 0$

So ${r}_{1} = - \frac{1}{2}$ and ${r}_{2} = 3$

Aug 8, 2018

$r = - \frac{1}{2} \text{ or } r = 3$

#### Explanation:

"rearrange into standard form ";ax^2+bx+c=0;a!=0

$\text{subtract "5r+3" from both sides}$

$2 {r}^{2} - 5 r - 3 = 0$

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

$\text{the factors of the product } 2 \times - 3 = - 6$

$\text{which sum to "-5" are "-6" and } + 1$

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

$2 {r}^{2} - 6 r + r - 3 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$\textcolor{red}{2 r} \left(r - 3\right) \textcolor{red}{+ 1} \left(r - 3\right) = 0$

$\text{take out the "color(blue)"common factor } \left(r - 3\right)$

$\left(r - 3\right) \left(\textcolor{red}{2 r + 1}\right) = 0$

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

$2 r + 1 = 0 \Rightarrow r = - \frac{1}{2}$

$r - 3 = 0 \Rightarrow r = 3$