.1x^2+55x+4000=0

Solution for .1x^2+55x+4000=0 equation:

.1x^2+55x+4000=0

a = .1; b = 55; c = +4000;
Δ = b2-4ac
Δ = 552-4·.1·4000
Δ = 1425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-5\sqrt{57}}{2*.1}=\frac{-55-5\sqrt{57}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+5\sqrt{57}}{2*.1}=\frac{-55+5\sqrt{57}}{0.2}
$