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

We get rid of parentheses

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