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4+2*1/2y=2y+4
We move all terms to the left:
4+2*1/2y-(2y+4)=0
Domain of the equation: 2y!=0We get rid of parentheses
y!=0/2
y!=0
y∈R
2*1/2y-2y-4+4=0
We multiply all the terms by the denominator
-2y*2y-4*2y+4*2y+2*1=0
We add all the numbers together, and all the variables
-2y*2y-4*2y+4*2y+2=0
Wy multiply elements
-4y^2-8y+8y+2=0
We add all the numbers together, and all the variables
-4y^2+2=0
a = -4; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-4)·2
Δ = 32
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-4}=\frac{0-4\sqrt{2}}{-8} =-\frac{4\sqrt{2}}{-8} =-\frac{\sqrt{2}}{-2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-4}=\frac{0+4\sqrt{2}}{-8} =\frac{4\sqrt{2}}{-8} =\frac{\sqrt{2}}{-2} $
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