7y^2+56y+40=0

Solution for 7y^2+56y+40=0 equation:

7y^2+56y+40=0

a = 7; b = 56; c = +40;
Δ = b2-4ac
Δ = 562-4·7·40
Δ = 2016
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:

$\sqrt{\Delta}=\sqrt{2016}=\sqrt{144*14}=\sqrt{144}*\sqrt{14}=12\sqrt{14}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-12\sqrt{14}}{2*7}=\frac{-56-12\sqrt{14}}{14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+12\sqrt{14}}{2*7}=\frac{-56+12\sqrt{14}}{14}
$