# How do you solve the quadratic 7x^2+5=-137 using any method?

Dec 29, 2016

There are no real solutions. But you can solve this using complex numbers. Then, the solutions (both imaginary) are:

${x}_{1} = + \sqrt{\frac{142}{7}} \cdot i \mathmr{and} {x}_{2} = - \sqrt{\frac{142}{7}} \cdot i$

#### Explanation:

You can probe there are no solutions in $\mathbb{R}$ clearing the $x$ as follows:

$7 {x}^{2} + 5 = - 137 \textcolor{w h i t e}{\text{." rArr color(white) ".}} 7 {x}^{2} = - 137 - 5 = - 142$,

then

$7 {x}^{2} = - 142 \textcolor{w h i t e}{\text{." rArr color(white) "." x^2 = - 142/7 color(white) "." rArr color(white) ".}} x = \pm \sqrt{- \frac{142}{7}} \notin \mathbb{R}$.

However, if we still want to give an answer, we can use complex numbers using the definition of the imaginary number $i$:

Since $i = \sqrt{- 1}$ we can write the solutions of the quadratic equation of the form:

$x = \pm \sqrt{- \frac{142}{7}} = \pm \sqrt{\frac{142}{7}} \cdot \sqrt{- 1} = \pm \sqrt{\frac{142}{7}} \cdot i \in \mathbb{C}$.