7(1-y)=-3y(y-2)

Solution for 7(1-y)=-3y(y-2) equation:

7(1-y)=-3y(y-2)

We move all terms to the left:

7(1-y)-(-3y(y-2))=0

We add all the numbers together, and all the variables

7(-1y+1)-(-3y(y-2))=0

We multiply parentheses

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


We calculate terms in parentheses: -(-3y(y-2)), so:

-3y(y-2)

We multiply parentheses

-3y^2+6y

Back to the equation:

-(-3y^2+6y)


We get rid of parentheses

3y^2-6y-7y+7=0

We add all the numbers together, and all the variables

3y^2-13y+7=0

a = 3; b = -13; c = +7;
Δ = b2-4ac
Δ = -132-4·3·7
Δ = 85
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{-(-13)-\sqrt{85}}{2*3}=\frac{13-\sqrt{85}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{85}}{2*3}=\frac{13+\sqrt{85}}{6}
$