-0.04x^2+300=0

Solution for -0.04x^2+300=0 equation:

-0.04x^2+300=0

a = -0.04; b = 0; c = +300;
Δ = b2-4ac
Δ = 02-4·(-0.04)·300
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-0.04}=\frac{0-4\sqrt{3}}{-0.08}
=-\frac{4\sqrt{3}}{-0.08}
=-\frac{\sqrt{3}}{-0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-0.04}=\frac{0+4\sqrt{3}}{-0.08}
=\frac{4\sqrt{3}}{-0.08}
=\frac{\sqrt{3}}{-0.02}
$