y2+10y-5=(y+3)(y-1)

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Solution for y2+10y-5=(y+3)(y-1) equation: 



y2+10y-5=(y+3)(y-1)

We move all terms to the left:

y2+10y-5-((y+3)(y-1))=0

We add all the numbers together, and all the variables

y^2+10y-((y+3)(y-1))-5=0

We multiply parentheses ..

y^2-((+y^2-1y+3y-3))+10y-5=0



We calculate terms in parentheses: -((+y^2-1y+3y-3)), so:

(+y^2-1y+3y-3)

We get rid of parentheses

y^2-1y+3y-3

We add all the numbers together, and all the variables

y^2+2y-3

Back to the equation:

-(y^2+2y-3)



We add all the numbers together, and all the variables

y^2+10y-(y^2+2y-3)-5=0

We get rid of parentheses

y^2-y^2+10y-2y+3-5=0

We add all the numbers together, and all the variables

8y-2=0

We move all terms containing y to the left, all other terms to the right

8y=2

y=2/8

y=1/4

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