18x+40=9x^2

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

18x+40=9x^2

We move all terms to the left:

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

determiningTheFunctionDomain
-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
$