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

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



(x-2)(x-2)+x(x-3)=4
We move all terms to the left:
(x-2)(x-2)+x(x-3)-(4)=0
We multiply parentheses
x^2+(x-2)(x-2)-3x-4=0
We multiply parentheses ..
x^2+(+x^2-2x-2x+4)-3x-4=0
We get rid of parentheses
x^2+x^2-2x-2x-3x+4-4=0
We add all the numbers together, and all the variables
2x^2-7x=0
a = 2; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2}=\frac{0}{4} =0 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2}=\frac{14}{4} =3+1/2 $

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