25k^2+20k-21=0

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

25k^2+20k-21=0

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

$\sqrt{\Delta}=\sqrt{2500}=50$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-50}{2*25}=\frac{-70}{50}
=-1+2/5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+50}{2*25}=\frac{30}{50}
=3/5
$