s(0)=-3(0-2/3)+19

Solution for s(0)=-3(0-2/3)+19 equation:

s(0)=-3(0-2/3)+19

We move all terms to the left:

s(0)-(-3(0-2/3)+19)=0

We add all the numbers together, and all the variables

s0-(-3(-2/3)+19)=0

We add all the numbers together, and all the variables

s-(-3(-2/3)+19)=0

We multiply all the terms by the denominator


s*3)+19)-(-3(-2=0

Wy multiply elements

3s^2-2=0

a = 3; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·3·(-2)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*3}=\frac{0-2\sqrt{6}}{6}
=-\frac{2\sqrt{6}}{6}
=-\frac{\sqrt{6}}{3}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*3}=\frac{0+2\sqrt{6}}{6}
=\frac{2\sqrt{6}}{6}
=\frac{\sqrt{6}}{3}
$