9x^2+18x=5

Solution for 9x^2+18x=5 equation:

9x^2+18x=5

We move all terms to the left:

9x^2+18x-(5)=0

a = 9; b = 18; c = -5;
Δ = b2-4ac
Δ = 182-4·9·(-5)
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{14}}{2*9}=\frac{-18-6\sqrt{14}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{14}}{2*9}=\frac{-18+6\sqrt{14}}{18}
$