(4x+3)(3x+9)=33

Solution for (4x+3)(3x+9)=33 equation:

(4x+3)(3x+9)=33

We move all terms to the left:

(4x+3)(3x+9)-(33)=0

We multiply parentheses ..

(+12x^2+36x+9x+27)-33=0

We get rid of parentheses

12x^2+36x+9x+27-33=0

We add all the numbers together, and all the variables

12x^2+45x-6=0

a = 12; b = 45; c = -6;
Δ = b2-4ac
Δ = 452-4·12·(-6)
Δ = 2313
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:

$\sqrt{\Delta}=\sqrt{2313}=\sqrt{9*257}=\sqrt{9}*\sqrt{257}=3\sqrt{257}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-3\sqrt{257}}{2*12}=\frac{-45-3\sqrt{257}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+3\sqrt{257}}{2*12}=\frac{-45+3\sqrt{257}}{24}
$