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1/2y=3y-5
We move all terms to the left:
1/2y-(3y-5)=0
Domain of the equation: 2y!=0We get rid of parentheses
y!=0/2
y!=0
y∈R
1/2y-3y+5=0
We multiply all the terms by the denominator
-3y*2y+5*2y+1=0
Wy multiply elements
-6y^2+10y+1=0
a = -6; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·(-6)·1
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{31}}{2*-6}=\frac{-10-2\sqrt{31}}{-12} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{31}}{2*-6}=\frac{-10+2\sqrt{31}}{-12} $
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