7y^2+16y-48=0

Solution for 7y^2+16y-48=0 equation:

7y^2+16y-48=0

a = 7; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·7·(-48)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-40}{2*7}=\frac{-56}{14}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+40}{2*7}=\frac{24}{14}
=1+5/7
$