5y^2+20y-35=0

Solution for 5y^2+20y-35=0 equation:

5y^2+20y-35=0

a = 5; b = 20; c = -35;
Δ = b2-4ac
Δ = 202-4·5·(-35)
Δ = 1100
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:

$\sqrt{\Delta}=\sqrt{1100}=\sqrt{100*11}=\sqrt{100}*\sqrt{11}=10\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{11}}{2*5}=\frac{-20-10\sqrt{11}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{11}}{2*5}=\frac{-20+10\sqrt{11}}{10}
$