# How do you solve 4x^2-4x=15?

Jun 4, 2018

${x}_{1} = \frac{5}{2}$ or ${x}_{2} = - \frac{3}{2}$

#### Explanation:

Writing your equation in the form
$4 {x}^{2} - 4 x - 15 = 0$
dividing by $4$
${x}^{2} - x - \frac{15}{4} = 0$
${x}_{1 , 2} = \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{15}{4}}$
so we get

${x}_{1} = \frac{5}{2}$

${x}_{2} = - \frac{3}{2}$

Jun 4, 2018

$x = - \frac{3}{2} \text{ or } x = \frac{5}{2}$

#### Explanation:

"rearrange in standard form ";ax^2+bx+c=0

$\text{subtract 15 from both sides}$

$4 {x}^{2} - 4 x - 15 = 0$

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

$\text{the factors of the product } 4 \times - 15 = - 60$

$\text{which sum to - 4 are + 6 and - 10}$

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

$4 {x}^{2} + 6 x - 10 x - 15 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$\textcolor{red}{2 x} \left(2 x + 3\right) \textcolor{red}{- 5} \left(2 x + 3\right) = 0$

$\text{take out the "color(blue)"common factor } \left(2 x + 3\right)$

$\left(2 x + 3\right) \left(\textcolor{red}{2 x - 5}\right) = 0$

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

$2 x + 3 = 0 \Rightarrow x = - \frac{3}{2}$

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