21k^2+15k-6=0

Solution for 21k^2+15k-6=0 equation:

21k^2+15k-6=0

a = 21; b = 15; c = -6;
Δ = b2-4ac
Δ = 152-4·21·(-6)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-27}{2*21}=\frac{-42}{42}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+27}{2*21}=\frac{12}{42}
=2/7
$