9x^2+18x=40

Solution for 9x^2+18x=40 equation:

9x^2+18x=40

We move all terms to the left:

9x^2+18x-(40)=0

a = 9; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·9·(-40)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*9}=\frac{-60}{18}
=-3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*9}=\frac{24}{18}
=1+1/3
$