(3+x)(3x-9)=48

Solution for (3+x)(3x-9)=48 equation:

(3+x)(3x-9)=48

We move all terms to the left:

(3+x)(3x-9)-(48)=0

We add all the numbers together, and all the variables

(x+3)(3x-9)-48=0

We multiply parentheses ..

(+3x^2-9x+9x-27)-48=0

We get rid of parentheses

3x^2-9x+9x-27-48=0

We add all the numbers together, and all the variables

3x^2-75=0

a = 3; b = 0; c = -75;
Δ = b2-4ac
Δ = 02-4·3·(-75)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*3}=\frac{-30}{6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*3}=\frac{30}{6}
=5
$