5y(2y+5)-2y(8y-5)=-4y(2y-9)-(y-50)

Solution for 5y(2y+5)-2y(8y-5)=-4y(2y-9)-(y-50) equation:

5y(2y+5)-2y(8y-5)=-4y(2y-9)-(y-50)

We move all terms to the left:

5y(2y+5)-2y(8y-5)-(-4y(2y-9)-(y-50))=0

We multiply parentheses

10y^2-16y^2+25y+10y-(-4y(2y-9)-(y-50))=0


We calculate terms in parentheses: -(-4y(2y-9)-(y-50)), so:

-4y(2y-9)-(y-50)

We multiply parentheses

-8y^2+36y-(y-50)

We get rid of parentheses

-8y^2+36y-y+50

We add all the numbers together, and all the variables

-8y^2+35y+50

Back to the equation:

-(-8y^2+35y+50)


We add all the numbers together, and all the variables

-6y^2-(-8y^2+35y+50)+35y=0

We get rid of parentheses

-6y^2+8y^2-35y+35y-50=0

We add all the numbers together, and all the variables

2y^2-50=0

a = 2; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·2·(-50)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*2}=\frac{-20}{4}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*2}=\frac{20}{4}
=5
$