19=s+s+(s2-5)

Solution for 19=s+s+(s2-5) equation:

19=s+s+(s2-5)

We move all terms to the left:

19-(s+s+(s2-5))=0

We add all the numbers together, and all the variables

-(s+s+(+s^2-5))+19=0


We calculate terms in parentheses: -(s+s+(+s^2-5)), so:

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

determiningTheFunctionDomain
(+s^2-5)+s+s

We add all the numbers together, and all the variables

(+s^2-5)+2s

We get rid of parentheses

s^2+2s-5

Back to the equation:

-(s^2+2s-5)


We get rid of parentheses

-s^2-2s+5+19=0

We add all the numbers together, and all the variables

-1s^2-2s+24=0

a = -1; b = -2; c = +24;
Δ = b2-4ac
Δ = -22-4·(-1)·24
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*-1}=\frac{-8}{-2}
=+4
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*-1}=\frac{12}{-2}
=-6
$