(5+35)(y^2+2y-3)=0

Solution for (5+35)(y^2+2y-3)=0 equation:

(5+35)(y^2+2y-3)=0

We add all the numbers together, and all the variables

40(y^2+2y-3)=0

We multiply parentheses

40y^2+80y-120=0

a = 40; b = 80; c = -120;
Δ = b2-4ac
Δ = 802-4·40·(-120)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-160}{2*40}=\frac{-240}{80}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+160}{2*40}=\frac{80}{80}
=1
$