3h^2-41h-14=0

Solution for 3h^2-41h-14=0 equation:

3h^2-41h-14=0

a = 3; b = -41; c = -14;
Δ = b2-4ac
Δ = -412-4·3·(-14)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-43}{2*3}=\frac{-2}{6}
=-1/3
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+43}{2*3}=\frac{84}{6}
=14
$