5y(y+4)=17+3y

Solution for 5y(y+4)=17+3y equation:

5y(y+4)=17+3y

We move all terms to the left:

5y(y+4)-(17+3y)=0

We add all the numbers together, and all the variables

5y(y+4)-(3y+17)=0

We multiply parentheses

5y^2+20y-(3y+17)=0

We get rid of parentheses

5y^2+20y-3y-17=0

We add all the numbers together, and all the variables

5y^2+17y-17=0

a = 5; b = 17; c = -17;
Δ = b2-4ac
Δ = 172-4·5·(-17)
Δ = 629
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{-(17)-\sqrt{629}}{2*5}=\frac{-17-\sqrt{629}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{629}}{2*5}=\frac{-17+\sqrt{629}}{10}
$