-3y^2+23y-30=0

Solution for -3y^2+23y-30=0 equation:

-3y^2+23y-30=0

a = -3; b = 23; c = -30;
Δ = b2-4ac
Δ = 232-4·(-3)·(-30)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-13}{2*-3}=\frac{-36}{-6}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+13}{2*-3}=\frac{-10}{-6}
=1+2/3
$