0.2x^2+10x-400=0

Solution for 0.2x^2+10x-400=0 equation:

0.2x^2+10x-400=0

a = 0.2; b = 10; c = -400;
Δ = b2-4ac
Δ = 102-4·0.2·(-400)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{105}}{2*0.2}=\frac{-10-2\sqrt{105}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{105}}{2*0.2}=\frac{-10+2\sqrt{105}}{0.4}
$