0.15x^2+8x+39=0

Solution for 0.15x^2+8x+39=0 equation:

0.15x^2+8x+39=0

a = 0.15; b = 8; c = +39;
Δ = b2-4ac
Δ = 82-4·0.15·39
Δ = 40.6
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{-(8)-\sqrt{40.6}}{2*0.15}=\frac{-8-\sqrt{40.6}}{0.3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+\sqrt{40.6}}{2*0.15}=\frac{-8+\sqrt{40.6}}{0.3}
$