-60q^2+150=0

Solution for -60q^2+150=0 equation:

-60q^2+150=0

a = -60; b = 0; c = +150;
Δ = b2-4ac
Δ = 02-4·(-60)·150
Δ = 36000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{36000}=\sqrt{3600*10}=\sqrt{3600}*\sqrt{10}=60\sqrt{10}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60\sqrt{10}}{2*-60}=\frac{0-60\sqrt{10}}{-120}
=-\frac{60\sqrt{10}}{-120}
=-\frac{\sqrt{10}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60\sqrt{10}}{2*-60}=\frac{0+60\sqrt{10}}{-120}
=\frac{60\sqrt{10}}{-120}
=\frac{\sqrt{10}}{-2}
$