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

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}
$