9d^2+d-9=0

Solution for 9d^2+d-9=0 equation:

9d^2+d-9=0

a = 9; b = 1; c = -9;
Δ = b2-4ac
Δ = 12-4·9·(-9)
Δ = 325
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5\sqrt{13}}{2*9}=\frac{-1-5\sqrt{13}}{18}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5\sqrt{13}}{2*9}=\frac{-1+5\sqrt{13}}{18}
$