# Solving equation?

## $6 {x}^{2} + x = 40$ Thank guys!

Dec 2, 2017

$2$ real solutions:

$x = 2.5 \setminus \quad , \setminus \quad x = - 2. \setminus \overline{6}$

#### Explanation:

Let’s first rearrange the equation:

$6 {x}^{2} + x - 40 = 0$

It’s a quadratic equation, which we can use the quadratic equation to solve.

Since the equation is in standard form, let’s define the variables:

• $a = 6$
• $b = 1$
• $c = - 40$

We can now plug those values into the quadratic formula:

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\setminus \implies x = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{{\left(1\right)}^{2} - 4 \left(6\right) \left(- 40\right)}}{2 \left(6\right)}$

$\setminus \implies x = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{1 + 960}}{12}$

$\setminus \implies x = \setminus \frac{- 1 \setminus \pm 31}{12}$

$\setminus \implies x = 2.5 \setminus \quad , \setminus \quad x = - 2. \setminus \overline{6}$