20n^2+42n-20=0

Solution for 20n^2+42n-20=0 equation:

20n^2+42n-20=0

a = 20; b = 42; c = -20;
Δ = b2-4ac
Δ = 422-4·20·(-20)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-58}{2*20}=\frac{-100}{40}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+58}{2*20}=\frac{16}{40}
=2/5
$