X(x+1)+(x-2)(x+3)=1

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

X(X+1)+(X-2)(X+3)=1

We move all terms to the left:

X(X+1)+(X-2)(X+3)-(1)=0

We multiply parentheses

X^2+X+(X-2)(X+3)-1=0

We multiply parentheses ..

X^2+(+X^2+3X-2X-6)+X-1=0

We get rid of parentheses

X^2+X^2+3X-2X+X-6-1=0

We add all the numbers together, and all the variables

2X^2+2X-7=0

a = 2; b = 2; c = -7;
Δ = b2-4ac
Δ = 22-4·2·(-7)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{15}}{2*2}=\frac{-2-2\sqrt{15}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{15}}{2*2}=\frac{-2+2\sqrt{15}}{4}
$