0.55x^2+20x-90=0

Solution for 0.55x^2+20x-90=0 equation:

0.55x^2+20x-90=0

a = 0.55; b = 20; c = -90;
Δ = b2-4ac
Δ = 202-4·0.55·(-90)
Δ = 598
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{-(20)-\sqrt{598}}{2*0.55}=\frac{-20-\sqrt{598}}{1.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{598}}{2*0.55}=\frac{-20+\sqrt{598}}{1.1}
$