9x+40=3x^2

Solution for 9x+40=3x^2 equation:

9x+40=3x^2

We move all terms to the left:

9x+40-(3x^2)=0

determiningTheFunctionDomain
-3x^2+9x+40=0

a = -3; b = 9; c = +40;
Δ = b2-4ac
Δ = 92-4·(-3)·40
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{561}}{2*-3}=\frac{-9-\sqrt{561}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{561}}{2*-3}=\frac{-9+\sqrt{561}}{-6}
$