(7-2y)/2-(2/5)(2-y)=11/4

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Solution for (7-2y)/2-(2/5)(2-y)=11/4 equation:



(7-2y)/2-(2/5)(2-y)=11/4
We move all terms to the left:
(7-2y)/2-(2/5)(2-y)-(11/4)=0
Domain of the equation: 5)(2-y)!=0
y∈R
We add all the numbers together, and all the variables
(-2y+7)/2-(+2/5)(-1y+2)-(+11/4)=0
We get rid of parentheses
(-2y+7)/2-(+2/5)(-1y+2)-11/4=0
We multiply parentheses ..
-(-2y^2+2/5*2)+(-2y+7)/2-11/4=0
We calculate fractions
(-(-2y^2+2*2*4)/()+(-8y)/()+()/()=0
We calculate terms in parentheses: +(-(-2y^2+2*2*4)/()+(-8y)/()+()/(), so:
-(-2y^2+2*2*4)/()+(-8y)/()+()/(
We add all the numbers together, and all the variables
-(-2y^2+2*2*4)/()+(-8y)/()+1
We multiply all the terms by the denominator
-(-2y^2+2*2*4)+(-8y)+1*()
We add all the numbers together, and all the variables
-(-2y^2+2*2*4)+(-8y)
We get rid of parentheses
2y^2-8y-2*2*4
We add all the numbers together, and all the variables
2y^2-8y-16
Back to the equation:
+(2y^2-8y-16)
We get rid of parentheses
2y^2-8y-16=0
a = 2; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·2·(-16)
Δ = 192
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:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{3}}{2*2}=\frac{8-8\sqrt{3}}{4} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{3}}{2*2}=\frac{8+8\sqrt{3}}{4} $

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