9k^2+10k-3=0

Solution for 9k^2+10k-3=0 equation:

9k^2+10k-3=0

a = 9; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·9·(-3)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{13}}{2*9}=\frac{-10-4\sqrt{13}}{18}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{13}}{2*9}=\frac{-10+4\sqrt{13}}{18}
$