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

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

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

We move all terms to the left:

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


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

5(y+1)2y-5

We multiply parentheses

10y^2+10y-5

Back to the equation:

-(10y^2+10y-5)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-10y^2-9y+5=0

a = -10; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·(-10)·5
Δ = 281
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{-(-9)-\sqrt{281}}{2*-10}=\frac{9-\sqrt{281}}{-20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{281}}{2*-10}=\frac{9+\sqrt{281}}{-20}
$