32k^2+21k-9=0

Solution for 32k^2+21k-9=0 equation:

32k^2+21k-9=0

a = 32; b = 21; c = -9;
Δ = b2-4ac
Δ = 212-4·32·(-9)
Δ = 1593
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1593}=\sqrt{9*177}=\sqrt{9}*\sqrt{177}=3\sqrt{177}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{177}}{2*32}=\frac{-21-3\sqrt{177}}{64}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{177}}{2*32}=\frac{-21+3\sqrt{177}}{64}
$