7(y-3)=2y(y-9)+2y

Solution for 7(y-3)=2y(y-9)+2y equation:

7(y-3)=2y(y-9)+2y

We move all terms to the left:

7(y-3)-(2y(y-9)+2y)=0

We multiply parentheses

7y-(2y(y-9)+2y)-21=0


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

2y(y-9)+2y

We add all the numbers together, and all the variables

2y+2y(y-9)

We multiply parentheses

2y^2+2y-18y

We add all the numbers together, and all the variables

2y^2-16y

Back to the equation:

-(2y^2-16y)


We get rid of parentheses

-2y^2+7y+16y-21=0

We add all the numbers together, and all the variables

-2y^2+23y-21=0

a = -2; b = 23; c = -21;
Δ = b2-4ac
Δ = 232-4·(-2)·(-21)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*-2}=\frac{-42}{-4}
=10+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*-2}=\frac{-4}{-4}
=1
$