4y(y-5)+5(5+2y)-10y=(2y-5)2

Solution for 4y(y-5)+5(5+2y)-10y=(2y-5)2 equation:

4y(y-5)+5(5+2y)-10y=(2y-5)2

We move all terms to the left:

4y(y-5)+5(5+2y)-10y-((2y-5)2)=0

We add all the numbers together, and all the variables

4y(y-5)+5(2y+5)-10y-((2y-5)2)=0

We add all the numbers together, and all the variables

-10y+4y(y-5)+5(2y+5)-((2y-5)2)=0

We multiply parentheses

4y^2-10y-20y+10y-((2y-5)2)+25=0


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

(2y-5)2

We multiply parentheses

4y-10

Back to the equation:

-(4y-10)


We add all the numbers together, and all the variables

4y^2-20y-(4y-10)+25=0

We get rid of parentheses

4y^2-20y-4y+10+25=0

We add all the numbers together, and all the variables

4y^2-24y+35=0

a = 4; b = -24; c = +35;
Δ = b2-4ac
Δ = -242-4·4·35
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4}{2*4}=\frac{20}{8}
=2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4}{2*4}=\frac{28}{8}
=3+1/2
$