7s(3s+1)=4(3+s)

Solution for 7s(3s+1)=4(3+s) equation:

7s(3s+1)=4(3+s)

We move all terms to the left:

7s(3s+1)-(4(3+s))=0

We add all the numbers together, and all the variables

7s(3s+1)-(4(s+3))=0

We multiply parentheses

21s^2+7s-(4(s+3))=0


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

4(s+3)

We multiply parentheses

4s+12

Back to the equation:

-(4s+12)


We get rid of parentheses

21s^2+7s-4s-12=0

We add all the numbers together, and all the variables

21s^2+3s-12=0

a = 21; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·21·(-12)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{113}}{2*21}=\frac{-3-3\sqrt{113}}{42}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{113}}{2*21}=\frac{-3+3\sqrt{113}}{42}
$