-5y^2+34y-48=0

Solution for -5y^2+34y-48=0 equation:

-5y^2+34y-48=0

a = -5; b = 34; c = -48;
Δ = b2-4ac
Δ = 342-4·(-5)·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-14}{2*-5}=\frac{-48}{-10}
=4+4/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+14}{2*-5}=\frac{-20}{-10}
=+2
$