# How do you solve 35x^ + 98x + 40=0?

Nov 11, 2015

Assuming the intended equation was $35 {x}^{2} + 98 x + 40 = 0$, use the quadratic formula to find:

$x = - \frac{7}{5} \pm \frac{\sqrt{1001}}{35}$

#### Explanation:

$35 {x}^{2} + 98 x + 40$ is of the form $a {x}^{2} + b x + c$ with $a = 35$, $b = 98$ and $c = 40$.

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 98 \pm \sqrt{{98}^{2} - \left(4 \cdot 35 \cdot 40\right)}}{2 \cdot 35} = - \frac{7}{5} \pm \frac{\sqrt{{49}^{2} - 1400}}{35}$

$= - \frac{7}{5} \pm \frac{\sqrt{1001}}{35}$