3x^2+43x+110=0

Solution for 3x^2+43x+110=0 equation:

3x^2+43x+110=0

a = 3; b = 43; c = +110;
Δ = b2-4ac
Δ = 432-4·3·110
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-23}{2*3}=\frac{-66}{6}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+23}{2*3}=\frac{-20}{6}
=-3+1/3
$