12d^2+19d+4=0

Solution for 12d^2+19d+4=0 equation:

12d^2+19d+4=0

a = 12; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·12·4
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*12}=\frac{-32}{24}
=-1+1/3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*12}=\frac{-6}{24}
=-1/4
$