# How do you solve 5x^2+98=0 using the quadratic formula?

Sep 11, 2016

You don't need the quadratic formula. But here goes:

Equation not valid in $\mathbb{R}$

#### Explanation:

$5 {x}^{2} + 98 = 0$

This is the same as:

$5 {x}^{2} + 0 x + 98 = 0$

For $a = 5$ $b = 0$ $c = 98$ :

Δ=b^2-4*a*c=0^2-4*5*98=-1960

Since Δ<0 the equation in impossible inside $\mathbb{R}$.

Simpler solution

$5 {x}^{2} + 98 = 0$

$5 {x}^{2} = - 98$

${x}^{2} = - \frac{98}{5}$

No real number can have a negative square, therefore the equation in impossible inside $\mathbb{R}$.