S=(5t-3)(6t+1)

Solution for S=(5t-3)(6t+1) equation:

=(5S-3)(6S+1)

We move all terms to the left:

-((5S-3)(6S+1))=0

We multiply parentheses ..

-((+30S^2+5S-18S-3))=0


We calculate terms in parentheses: -((+30S^2+5S-18S-3)), so:

(+30S^2+5S-18S-3)

We get rid of parentheses

30S^2+5S-18S-3

We add all the numbers together, and all the variables

30S^2-13S-3

Back to the equation:

-(30S^2-13S-3)


We get rid of parentheses

-30S^2+13S+3=0

a = -30; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·(-30)·3
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*-30}=\frac{-36}{-60}
=3/5
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*-30}=\frac{10}{-60}
=-1/6
$