-3x2+50=0

Solution for -3x2+50=0 equation:

-3x^2+50=0

a = -3; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-3)·50
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{6}}{2*-3}=\frac{0-10\sqrt{6}}{-6}
=-\frac{10\sqrt{6}}{-6}
=-\frac{5\sqrt{6}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{6}}{2*-3}=\frac{0+10\sqrt{6}}{-6}
=\frac{10\sqrt{6}}{-6}
=\frac{5\sqrt{6}}{-3}
$