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

Apr 5, 2016

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

#### Explanation:

Given:$\text{ } 2 {x}^{2} + 5 x = 3$
Write as:$\text{ } 2 {x}^{2} + 5 x - 3 = 0$

The whole number factors of 3 are 1 and 3.
The whole number factors of 2 are 1 and 2.

Attempt 1

$\left(2 x + 3\right) \left(x + 1\right) = 2 {x}^{2} + 2 x + 3 x + 3 \textcolor{red}{\text{ Fail}}$

Attempt 2

$\left(2 x + 1\right) \left(x - 3\right) = 2 {x}^{2} - 6 x + x - 3 \textcolor{red}{\text{ Fail}}$

Looks as though there is at least 1 non whole number solution to this so use the formula

$y = a {x}^{2} + b x + c \text{ where } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(2\right) \left(- 3\right)}}{2 \left(2\right)}$

$x = \frac{- 5 \pm \sqrt{25 + 24}}{4}$

$x = \frac{- 5 \pm \sqrt{49}}{4}$

$x = \frac{- 5 \pm 7}{4}$

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