2(s1+3)=s1(4s-5)

Solution for 2(s1+3)=s1(4s-5) equation:

2(s1+3)=s1(4s-5)

We move all terms to the left:

2(s1+3)-(s1(4s-5))=0

We add all the numbers together, and all the variables

2(s+3)-(s1(4s-5))=0

We multiply parentheses

2s-(s1(4s-5))+6=0


We calculate terms in parentheses: -(s1(4s-5)), so:

s1(4s-5)

We multiply parentheses

4s^2-5s

Back to the equation:

-(4s^2-5s)


We get rid of parentheses

-4s^2+2s+5s+6=0

We add all the numbers together, and all the variables

-4s^2+7s+6=0

a = -4; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·(-4)·6
Δ = 145
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{-(7)-\sqrt{145}}{2*-4}=\frac{-7-\sqrt{145}}{-8}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{145}}{2*-4}=\frac{-7+\sqrt{145}}{-8}
$