0.2x^2+0.8x-0.75=0

Solution for 0.2x^2+0.8x-0.75=0 equation:

0.2x^2+0.8x-0.75=0

a = 0.2; b = 0.8; c = -0.75;
Δ = b2-4ac
Δ = 0.82-4·0.2·(-0.75)
Δ = 1.24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.8)-\sqrt{1.24}}{2*0.2}=\frac{-0.8-\sqrt{1.24}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{1.24}}{2*0.2}=\frac{-0.8+\sqrt{1.24}}{0.4}
$