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

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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} $

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