s^2=7+s(1-3s)

Solution for s^2=7+s(1-3s) equation:

s^2=7+s(1-3s)

We move all terms to the left:

s^2-(7+s(1-3s))=0

We add all the numbers together, and all the variables

s^2-(7+s(-3s+1))=0


We calculate terms in parentheses: -(7+s(-3s+1)), so:

7+s(-3s+1)

determiningTheFunctionDomain
s(-3s+1)+7

We multiply parentheses

-3s^2+s+7

Back to the equation:

-(-3s^2+s+7)


We get rid of parentheses

s^2+3s^2-s-7=0

We add all the numbers together, and all the variables

4s^2-1s-7=0

a = 4; b = -1; c = -7;
Δ = b2-4ac
Δ = -12-4·4·(-7)
Δ = 113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{113}}{2*4}=\frac{1-\sqrt{113}}{8}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{113}}{2*4}=\frac{1+\sqrt{113}}{8}
$