(3x^2)+40x+44=0

Solution for (3x^2)+40x+44=0 equation:

(3x^2)+40x+44=0

a = 3; b = 40; c = +44;
Δ = b2-4ac
Δ = 402-4·3·44
Δ = 1072
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:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{67}}{2*3}=\frac{-40-4\sqrt{67}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{67}}{2*3}=\frac{-40+4\sqrt{67}}{6}
$