# How do you solve sqrt(58n) = 125(n + 4) ?

Nov 9, 2015

You need to square the both parts.
As $\sqrt{58 n}$ is only defined for $n > 0$, both parts of your equation are positive, so you can do so safely.

$\sqrt{58 n} = 125 \left(n + 4\right)$
$58 n = {125}^{2} {\left(n + 4\right)}^{2}$
$58 n = {125}^{2} \left({n}^{2} + 8 n + 16\right)$

Dividing both sides of the equation by ${125}^{2}$ yields:
$\frac{58}{125} ^ 2 n = {n}^{2} + 8 n + 16$
$\iff$
${n}^{2} + \left(8 - \frac{58}{125} ^ 2\right) n + 16 = 0$

This is a regular quadratic equation which can be solved by regular means. The numbers are really ugly though..