4s-4=3s(s-3)+5+s

Solution for 4s-4=3s(s-3)+5+s equation:

4s-4=3s(s-3)+5+s

We move all terms to the left:

4s-4-(3s(s-3)+5+s)=0


We calculate terms in parentheses: -(3s(s-3)+5+s), so:

3s(s-3)+5+s

determiningTheFunctionDomain
3s(s-3)+s+5

We add all the numbers together, and all the variables

s+3s(s-3)+5

We multiply parentheses

3s^2+s-9s+5

We add all the numbers together, and all the variables

3s^2-8s+5

Back to the equation:

-(3s^2-8s+5)


We get rid of parentheses

-3s^2+4s+8s-5-4=0

We add all the numbers together, and all the variables

-3s^2+12s-9=0

a = -3; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·(-3)·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*-3}=\frac{-18}{-6}
=+3
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*-3}=\frac{-6}{-6}
=1
$