(x-3)(x+4)=(x+2)

Solution for (x-3)(x+4)=(x+2) equation:

(x-3)(x+4)=(x+2)

We move all terms to the left:

(x-3)(x+4)-((x+2))=0

We multiply parentheses ..

(+x^2+4x-3x-12)-((x+2))=0


We calculate terms in parentheses: -((x+2)), so:

(x+2)

We get rid of parentheses

x+2

Back to the equation:

-(x+2)


We get rid of parentheses

x^2+4x-3x-x-12-2=0

We add all the numbers together, and all the variables

x^2-14=0

a = 1; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·1·(-14)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*1}=\frac{0-2\sqrt{14}}{2}
=-\frac{2\sqrt{14}}{2}
=-\sqrt{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*1}=\frac{0+2\sqrt{14}}{2}
=\frac{2\sqrt{14}}{2}
=\sqrt{14}
$