-2y^2+15y-28=0

Solution for -2y^2+15y-28=0 equation:

-2y^2+15y-28=0

a = -2; b = 15; c = -28;
Δ = b2-4ac
Δ = 152-4·(-2)·(-28)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-1}{2*-2}=\frac{-16}{-4}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+1}{2*-2}=\frac{-14}{-4}
=3+1/2
$