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3/2x+3=x+1
We move all terms to the left:
3/2x+3-(x+1)=0
Domain of the equation: 2x!=0We get rid of parentheses
x!=0/2
x!=0
x∈R
3/2x-x-1+3=0
We multiply all the terms by the denominator
-x*2x-1*2x+3*2x+3=0
Wy multiply elements
-2x^2-2x+6x+3=0
We add all the numbers together, and all the variables
-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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