9k2+18k-43=0

Solution for 9k2+18k-43=0 equation:

9k^2+18k-43=0

a = 9; b = 18; c = -43;
Δ = b2-4ac
Δ = 182-4·9·(-43)
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12\sqrt{13}}{2*9}=\frac{-18-12\sqrt{13}}{18}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12\sqrt{13}}{2*9}=\frac{-18+12\sqrt{13}}{18}
$