28=(6x+1)(3x+7)

Solution for 28=(6x+1)(3x+7) equation:

28=(6x+1)(3x+7)

We move all terms to the left:

28-((6x+1)(3x+7))=0

We multiply parentheses ..

-((+18x^2+42x+3x+7))+28=0


We calculate terms in parentheses: -((+18x^2+42x+3x+7)), so:

(+18x^2+42x+3x+7)

We get rid of parentheses

18x^2+42x+3x+7

We add all the numbers together, and all the variables

18x^2+45x+7

Back to the equation:

-(18x^2+45x+7)


We get rid of parentheses

-18x^2-45x-7+28=0

We add all the numbers together, and all the variables

-18x^2-45x+21=0

a = -18; b = -45; c = +21;
Δ = b2-4ac
Δ = -452-4·(-18)·21
Δ = 3537
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3537}=\sqrt{9*393}=\sqrt{9}*\sqrt{393}=3\sqrt{393}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-3\sqrt{393}}{2*-18}=\frac{45-3\sqrt{393}}{-36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+3\sqrt{393}}{2*-18}=\frac{45+3\sqrt{393}}{-36}
$