3y^2+5y-28=0

Solution for 3y^2+5y-28=0 equation:

3y^2+5y-28=0

a = 3; b = 5; c = -28;
Δ = b2-4ac
Δ = 52-4·3·(-28)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-19}{2*3}=\frac{-24}{6}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+19}{2*3}=\frac{14}{6}
=2+1/3
$