x^2+.50x=1.1*10^-10

Solution for x^2+.50x=1.1*10^-10 equation:

x^2+.50x=1.1*10^-10

We move all terms to the left:

x^2+.50x-(1.1*10^-10)=0

We add all the numbers together, and all the variables

x^2+.50x-10-1.1E=0

We add all the numbers together, and all the variables

x^2+.50x-12.990110011305=0

a = 1; b = .50; c = -12.990110011305;
Δ = b2-4ac
Δ = .502-4·1·(-12.990110011305)
Δ = 52.21044004522
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{-(.50)-\sqrt{52.21044004522}}{2*1}=\frac{-0.5-\sqrt{52.21044004522}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.50)+\sqrt{52.21044004522}}{2*1}=\frac{-0.5+\sqrt{52.21044004522}}{2}
$