21n^2+20n-25=0

Solution for 21n^2+20n-25=0 equation:

21n^2+20n-25=0

a = 21; b = 20; c = -25;
Δ = b2-4ac
Δ = 202-4·21·(-25)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-50}{2*21}=\frac{-70}{42}
=-1+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+50}{2*21}=\frac{30}{42}
=5/7
$