4(2s-3)=s(s+1)

Solution for 4(2s-3)=s(s+1) equation:

4(2s-3)=s(s+1)

We move all terms to the left:

4(2s-3)-(s(s+1))=0

We multiply parentheses

8s-(s(s+1))-12=0


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

s(s+1)

We multiply parentheses

s^2+s

Back to the equation:

-(s^2+s)


We get rid of parentheses

-s^2+8s-s-12=0

We add all the numbers together, and all the variables

-1s^2+7s-12=0

a = -1; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·(-1)·(-12)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*-1}=\frac{-8}{-2}
=+4
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*-1}=\frac{-6}{-2}
=+3
$