3(y-10)+1=y(y-8)

Solution for 3(y-10)+1=y(y-8) equation:

3(y-10)+1=y(y-8)

We move all terms to the left:

3(y-10)+1-(y(y-8))=0

We multiply parentheses

3y-(y(y-8))-30+1=0


We calculate terms in parentheses: -(y(y-8)), so:

y(y-8)

We multiply parentheses

y^2-8y

Back to the equation:

-(y^2-8y)


We add all the numbers together, and all the variables

3y-(y^2-8y)-29=0

We get rid of parentheses

-y^2+3y+8y-29=0

We add all the numbers together, and all the variables

-1y^2+11y-29=0

a = -1; b = 11; c = -29;
Δ = b2-4ac
Δ = 112-4·(-1)·(-29)
Δ = 5
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{-(11)-\sqrt{5}}{2*-1}=\frac{-11-\sqrt{5}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{5}}{2*-1}=\frac{-11+\sqrt{5}}{-2}
$