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-1/2y+12=-0.5+12y
We move all terms to the left:
-1/2y+12-(-0.5+12y)=0
Domain of the equation: 2y!=0We add all the numbers together, and all the variables
y!=0/2
y!=0
y∈R
-1/2y-(12y-0.5)+12=0
We get rid of parentheses
-1/2y-12y+0.5+12=0
We multiply all the terms by the denominator
-12y*2y+(0.5)*2y+12*2y-1=0
We multiply parentheses
-12y*2y+y+12*2y-1=0
Wy multiply elements
-24y^2+y+24y-1=0
We add all the numbers together, and all the variables
-24y^2+25y-1=0
a = -24; b = 25; c = -1;
Δ = b2-4ac
Δ = 252-4·(-24)·(-1)
Δ = 529
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*-24}=\frac{-48}{-48} =1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*-24}=\frac{-2}{-48} =1/24 $
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