.02x^2-8x+600=0

Solution for .02x^2-8x+600=0 equation:

.02x^2-8x+600=0

a = .02; b = -8; c = +600;
Δ = b2-4ac
Δ = -82-4·.02·600
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*.02}=\frac{4}{0.04}
=100
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*.02}=\frac{12}{0.04}
=300
$