8y^2+40y=48

Solution for 8y^2+40y=48 equation:

8y^2+40y=48

We move all terms to the left:

8y^2+40y-(48)=0

a = 8; b = 40; c = -48;
Δ = b2-4ac
Δ = 402-4·8·(-48)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-56}{2*8}=\frac{-96}{16}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+56}{2*8}=\frac{16}{16}
=1
$