3y^2+43y+14=0

Solution for 3y^2+43y+14=0 equation:

3y^2+43y+14=0

a = 3; b = 43; c = +14;
Δ = b2-4ac
Δ = 432-4·3·14
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-41}{2*3}=\frac{-84}{6}
=-14
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+41}{2*3}=\frac{-2}{6}
=-1/3
$