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X=3/2X+2=7/2
We move all terms to the left:
X-(3/2X+2)=0
Domain of the equation: 2X+2)!=0We get rid of parentheses
X∈R
X-3/2X-2=0
We multiply all the terms by the denominator
X*2X-2*2X-3=0
Wy multiply elements
2X^2-4X-3=0
a = 2; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·2·(-3)
Δ = 40
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*2}=\frac{4-2\sqrt{10}}{4} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*2}=\frac{4+2\sqrt{10}}{4} $
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