X(x+1)+(x+12)(x+3)=42

Solution for X(x+1)+(x+12)(x+3)=42 equation:

X(X+1)+(X+12)(X+3)=42

We move all terms to the left:

X(X+1)+(X+12)(X+3)-(42)=0

We multiply parentheses

X^2+X+(X+12)(X+3)-42=0

We multiply parentheses ..

X^2+(+X^2+3X+12X+36)+X-42=0

We get rid of parentheses

X^2+X^2+3X+12X+X+36-42=0

We add all the numbers together, and all the variables

2X^2+16X-6=0

a = 2; b = 16; c = -6;
Δ = b2-4ac
Δ = 162-4·2·(-6)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{19}}{2*2}=\frac{-16-4\sqrt{19}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{19}}{2*2}=\frac{-16+4\sqrt{19}}{4}
$