24y^2+50y+19=0

Solution for 24y^2+50y+19=0 equation:

24y^2+50y+19=0

a = 24; b = 50; c = +19;
Δ = b2-4ac
Δ = 502-4·24·19
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-26}{2*24}=\frac{-76}{48}
=-1+7/12
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+26}{2*24}=\frac{-24}{48}
=-1/2
$