21k^2+23k-20=0

Solution for 21k^2+23k-20=0 equation:

21k^2+23k-20=0

a = 21; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·21·(-20)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-47}{2*21}=\frac{-70}{42}
=-1+2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+47}{2*21}=\frac{24}{42}
=4/7
$