5y(+2)=7(y+1)+1

Solution for 5y(+2)=7(y+1)+1 equation:

5y(+2)=7(y+1)+1

We move all terms to the left:

5y(+2)-(7(y+1)+1)=0

We add all the numbers together, and all the variables

5y2-(7(y+1)+1)=0

We add all the numbers together, and all the variables

5y^2-(7(y+1)+1)=0


We calculate terms in parentheses: -(7(y+1)+1), so:

7(y+1)+1

We multiply parentheses

7y+7+1

We add all the numbers together, and all the variables

7y+8

Back to the equation:

-(7y+8)


We get rid of parentheses

5y^2-7y-8=0

a = 5; b = -7; c = -8;
Δ = b2-4ac
Δ = -72-4·5·(-8)
Δ = 209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{209}}{2*5}=\frac{7-\sqrt{209}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{209}}{2*5}=\frac{7+\sqrt{209}}{10}
$