5y+2=(1/2)(10y+4)

Solution for 5y+2=(1/2)(10y+4) equation:

5y+2=(1/2)(10y+4)

We move all terms to the left:

5y+2-((1/2)(10y+4))=0


Domain of the equation: 2)(10y+4))!=0

y∈R


We add all the numbers together, and all the variables

5y-((+1/2)(10y+4))+2=0

We multiply parentheses ..

-((+10y^2+1/2*4))+5y+2=0

We multiply all the terms by the denominator


-((+10y^2+1+5y*2*4))+2*2*4))=0


We calculate terms in parentheses: -((+10y^2+1+5y*2*4)), so:

(+10y^2+1+5y*2*4)

We get rid of parentheses

10y^2+5y*2*4+1

Wy multiply elements

10y^2+40y*4+1

Wy multiply elements

10y^2+160y+1

Back to the equation:

-(10y^2+160y+1)


We add all the numbers together, and all the variables

-(10y^2+160y+1)=0

We get rid of parentheses

-10y^2-160y-1=0

a = -10; b = -160; c = -1;
Δ = b2-4ac
Δ = -1602-4·(-10)·(-1)
Δ = 25560
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:

$\sqrt{\Delta}=\sqrt{25560}=\sqrt{36*710}=\sqrt{36}*\sqrt{710}=6\sqrt{710}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-6\sqrt{710}}{2*-10}=\frac{160-6\sqrt{710}}{-20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+6\sqrt{710}}{2*-10}=\frac{160+6\sqrt{710}}{-20}
$