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

Solution for -7y^2-40y+12=0 equation:

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

a = -7; b = -40; c = +12;
Δ = b2-4ac
Δ = -402-4·(-7)·12
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-44}{2*-7}=\frac{-4}{-14}
=2/7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+44}{2*-7}=\frac{84}{-14}
=-6
$