0,05x^2+1.1x-250=0

Solution for 0,05x^2+1.1x-250=0 equation:

0.05x^2+1.1x-250=0

a = 0.05; b = 1.1; c = -250;
Δ = b2-4ac
Δ = 1.12-4·0.05·(-250)
Δ = 51.21
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{-(1.1)-\sqrt{51.21}}{2*0.05}=\frac{-1.1-\sqrt{51.21}}{0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{51.21}}{2*0.05}=\frac{-1.1+\sqrt{51.21}}{0.1}
$