(3x+3)(x+3)=x^2+105

Solution for (3x+3)(x+3)=x^2+105 equation:

(3x+3)(x+3)=x^2+105

We move all terms to the left:

(3x+3)(x+3)-(x^2+105)=0

We get rid of parentheses

-x^2+(3x+3)(x+3)-105=0

We multiply parentheses ..

-x^2+(+3x^2+9x+3x+9)-105=0

We add all the numbers together, and all the variables

-1x^2+(+3x^2+9x+3x+9)-105=0

We get rid of parentheses

-1x^2+3x^2+9x+3x+9-105=0

We add all the numbers together, and all the variables

2x^2+12x-96=0

a = 2; b = 12; c = -96;
Δ = b2-4ac
Δ = 122-4·2·(-96)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{57}}{2*2}=\frac{-12-4\sqrt{57}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{57}}{2*2}=\frac{-12+4\sqrt{57}}{4}
$