9x+3x^2=27

Solution for 9x+3x^2=27 equation:

9x+3x^2=27

We move all terms to the left:

9x+3x^2-(27)=0

a = 3; b = 9; c = -27;
Δ = b2-4ac
Δ = 92-4·3·(-27)
Δ = 405
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:

$\sqrt{\Delta}=\sqrt{405}=\sqrt{81*5}=\sqrt{81}*\sqrt{5}=9\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9\sqrt{5}}{2*3}=\frac{-9-9\sqrt{5}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9\sqrt{5}}{2*3}=\frac{-9+9\sqrt{5}}{6}
$