(3s+1)(s-4)=5(s-1)

Solution for (3s+1)(s-4)=5(s-1) equation:

(3s+1)(s-4)=5(s-1)

We move all terms to the left:

(3s+1)(s-4)-(5(s-1))=0

We multiply parentheses ..

(+3s^2-12s+s-4)-(5(s-1))=0


We calculate terms in parentheses: -(5(s-1)), so:

5(s-1)

We multiply parentheses

5s-5

Back to the equation:

-(5s-5)


We get rid of parentheses

3s^2-12s+s-5s-4+5=0

We add all the numbers together, and all the variables

3s^2-16s+1=0

a = 3; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·3·1
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{61}}{2*3}=\frac{16-2\sqrt{61}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{61}}{2*3}=\frac{16+2\sqrt{61}}{6}
$