1.1x^2-20x+50=0

Solution for 1.1x^2-20x+50=0 equation:

1.1x^2-20x+50=0

a = 1.1; b = -20; c = +50;
Δ = b2-4ac
Δ = -202-4·1.1·50
Δ = 180
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{180}}{2*1.1}=\frac{20-\sqrt{180}}{2.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{180}}{2*1.1}=\frac{20+\sqrt{180}}{2.2}
$