(3x-2)(x+6)-(2x)(x+1)=103

Solution for (3x-2)(x+6)-(2x)(x+1)=103 equation:

(3x-2)(x+6)-(2x)(x+1)=103

We move all terms to the left:

(3x-2)(x+6)-(2x)(x+1)-(103)=0

We multiply parentheses

-2x^2+(3x-2)(x+6)-2x-103=0

We multiply parentheses ..

-2x^2+(+3x^2+18x-2x-12)-2x-103=0

We get rid of parentheses

-2x^2+3x^2+18x-2x-2x-12-103=0

We add all the numbers together, and all the variables

x^2+14x-115=0

a = 1; b = 14; c = -115;
Δ = b2-4ac
Δ = 142-4·1·(-115)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{41}}{2*1}=\frac{-14-4\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{41}}{2*1}=\frac{-14+4\sqrt{41}}{2}
$