12y^2+40y=-7

Solution for 12y^2+40y=-7 equation:

12y^2+40y=-7

We move all terms to the left:

12y^2+40y-(-7)=0

We add all the numbers together, and all the variables

12y^2+40y+7=0

a = 12; b = 40; c = +7;
Δ = b2-4ac
Δ = 402-4·12·7
Δ = 1264
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{79}}{2*12}=\frac{-40-4\sqrt{79}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{79}}{2*12}=\frac{-40+4\sqrt{79}}{24}
$