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

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



(x+3)(x-2)=(4-x)2
We move all terms to the left:
(x+3)(x-2)-((4-x)2)=0
We add all the numbers together, and all the variables
(x+3)(x-2)-((-1x+4)2)=0
We multiply parentheses ..
(+x^2-2x+3x-6)-((-1x+4)2)=0
We calculate terms in parentheses: -((-1x+4)2), so:
(-1x+4)2
We multiply parentheses
-2x+8
Back to the equation:
-(-2x+8)
We get rid of parentheses
x^2-2x+3x+2x-6-8=0
We add all the numbers together, and all the variables
x^2+3x-14=0
a = 1; b = 3; c = -14;
Δ = b2-4ac
Δ = 32-4·1·(-14)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{65}}{2*1}=\frac{-3-\sqrt{65}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{65}}{2*1}=\frac{-3+\sqrt{65}}{2} $

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