0.2x^2+100x-1500=0

Solution for 0.2x^2+100x-1500=0 equation:

0.2x^2+100x-1500=0

a = 0.2; b = 100; c = -1500;
Δ = b2-4ac
Δ = 1002-4·0.2·(-1500)
Δ = 11200
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:

$\sqrt{\Delta}=\sqrt{11200}=\sqrt{1600*7}=\sqrt{1600}*\sqrt{7}=40\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-40\sqrt{7}}{2*0.2}=\frac{-100-40\sqrt{7}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+40\sqrt{7}}{2*0.2}=\frac{-100+40\sqrt{7}}{0.4}
$