3d^2+19d-14=0

Solution for 3d^2+19d-14=0 equation:

3d^2+19d-14=0

a = 3; b = 19; c = -14;
Δ = b2-4ac
Δ = 192-4·3·(-14)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-23}{2*3}=\frac{-42}{6}
=-7
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*3}=\frac{4}{6}
=2/3
$