0.03x^2+20x+165=0

Solution for 0.03x^2+20x+165=0 equation:

0.03x^2+20x+165=0

a = 0.03; b = 20; c = +165;
Δ = b2-4ac
Δ = 202-4·0.03·165
Δ = 380.2
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{380.2}}{2*0.03}=\frac{-20-\sqrt{380.2}}{0.06}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{380.2}}{2*0.03}=\frac{-20+\sqrt{380.2}}{0.06}
$