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(x-3/2x+1)=4
We move all terms to the left:
(x-3/2x+1)-(4)=0
Domain of the equation: 2x+1)!=0We get rid of parentheses
x∈R
x-3/2x+1-4=0
We multiply all the terms by the denominator
x*2x+1*2x-4*2x-3=0
Wy multiply elements
2x^2+2x-8x-3=0
We add all the numbers together, and all the variables
2x^2-6x-3=0
a = 2; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·2·(-3)
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*2}=\frac{6-2\sqrt{15}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*2}=\frac{6+2\sqrt{15}}{4} $
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