20y(y+2)=4(4y+5)

Solution for 20y(y+2)=4(4y+5) equation:

20y(y+2)=4(4y+5)

We move all terms to the left:

20y(y+2)-(4(4y+5))=0

We multiply parentheses

20y^2+40y-(4(4y+5))=0


We calculate terms in parentheses: -(4(4y+5)), so:

4(4y+5)

We multiply parentheses

16y+20

Back to the equation:

-(16y+20)


We get rid of parentheses

20y^2+40y-16y-20=0

We add all the numbers together, and all the variables

20y^2+24y-20=0

a = 20; b = 24; c = -20;
Δ = b2-4ac
Δ = 242-4·20·(-20)
Δ = 2176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{34}}{2*20}=\frac{-24-8\sqrt{34}}{40}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{34}}{2*20}=\frac{-24+8\sqrt{34}}{40}
$