-3x^2+150=0

Solution for -3x^2+150=0 equation:

-3x^2+150=0

a = -3; b = 0; c = +150;
Δ = b2-4ac
Δ = 02-4·(-3)·150
Δ = 1800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{2}}{2*-3}=\frac{0-30\sqrt{2}}{-6}
=-\frac{30\sqrt{2}}{-6}
=-\frac{5\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{2}}{2*-3}=\frac{0+30\sqrt{2}}{-6}
=\frac{30\sqrt{2}}{-6}
=\frac{5\sqrt{2}}{-1}
$