20n2+2n-35=0

Solution for 20n2+2n-35=0 equation:

20n^2+2n-35=0

a = 20; b = 2; c = -35;
Δ = b2-4ac
Δ = 22-4·20·(-35)
Δ = 2804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2804}=\sqrt{4*701}=\sqrt{4}*\sqrt{701}=2\sqrt{701}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{701}}{2*20}=\frac{-2-2\sqrt{701}}{40}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{701}}{2*20}=\frac{-2+2\sqrt{701}}{40}
$