5y(y+1)=20(y+1)

Solution for 5y(y+1)=20(y+1) equation:

5y(y+1)=20(y+1)

We move all terms to the left:

5y(y+1)-(20(y+1))=0

We multiply parentheses

5y^2+5y-(20(y+1))=0


We calculate terms in parentheses: -(20(y+1)), so:

20(y+1)

We multiply parentheses

20y+20

Back to the equation:

-(20y+20)


We get rid of parentheses

5y^2+5y-20y-20=0

We add all the numbers together, and all the variables

5y^2-15y-20=0

a = 5; b = -15; c = -20;
Δ = b2-4ac
Δ = -152-4·5·(-20)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-25}{2*5}=\frac{-10}{10}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+25}{2*5}=\frac{40}{10}
=4
$