0.25x^2+11x-20=0

Solution for 0.25x^2+11x-20=0 equation:

0.25x^2+11x-20=0

a = 0.25; b = 11; c = -20;
Δ = b2-4ac
Δ = 112-4·0.25·(-20)
Δ = 141
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{-(11)-\sqrt{141}}{2*0.25}=\frac{-11-\sqrt{141}}{0.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{141}}{2*0.25}=\frac{-11+\sqrt{141}}{0.5}
$