(4x+9)x=99

Solution for (4x+9)x=99 equation:

(4x+9)x=99

We move all terms to the left:

(4x+9)x-(99)=0

We multiply parentheses

4x^2+9x-99=0

a = 4; b = 9; c = -99;
Δ = b2-4ac
Δ = 92-4·4·(-99)
Δ = 1665
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:

$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{185}}{2*4}=\frac{-9-3\sqrt{185}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{185}}{2*4}=\frac{-9+3\sqrt{185}}{8}
$