2x^2+18x+39=0

Solution for 2x^2+18x+39=0 equation:

2x^2+18x+39=0

a = 2; b = 18; c = +39;
Δ = b2-4ac
Δ = 182-4·2·39
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{3}}{2*2}=\frac{-18-2\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{3}}{2*2}=\frac{-18+2\sqrt{3}}{4}
$