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

Aug 16, 2015

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{41}}{4}$

#### Explanation:

You can solve this quadratic equation by using the quadratic formula, which tells you that for any general form quadratic equation

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

the two roots of the equation take the form

color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)

So, start by adding $- 5$ to both sides of the equation to get

$2 {x}^{2} + x - 5 = \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}$

$2 {x}^{2} + x - 5 = 0$

Notice that you have $a = 2$, $b = 1$, and $c = - 5$. This means that the two solutions will be

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 2 \cdot \left(- 5\right)}}{2 \cdot 2}$

${x}_{1 , 2} = \textcolor{g r e e n}{\frac{- 1 \pm \sqrt{41}}{4}}$

You can simplify this if you want to get

${x}_{1} = \frac{- 1 + \sqrt{41}}{4} \cong 1.35078$

and

${x}_{2} = \frac{- 1 - \sqrt{41}}{4} \cong - 1.85078$