3+6s=4s(s-3)+19

Solution for 3+6s=4s(s-3)+19 equation:

3+6s=4s(s-3)+19

We move all terms to the left:

3+6s-(4s(s-3)+19)=0


We calculate terms in parentheses: -(4s(s-3)+19), so:

4s(s-3)+19

We multiply parentheses

4s^2-12s+19

Back to the equation:

-(4s^2-12s+19)


We get rid of parentheses

-4s^2+6s+12s-19+3=0

We add all the numbers together, and all the variables

-4s^2+18s-16=0

a = -4; b = 18; c = -16;
Δ = b2-4ac
Δ = 182-4·(-4)·(-16)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{17}}{2*-4}=\frac{-18-2\sqrt{17}}{-8}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{17}}{2*-4}=\frac{-18+2\sqrt{17}}{-8}
$