42=(2x+1)(x+3)

Solution for 42=(2x+1)(x+3) equation:

42=(2x+1)(x+3)

We move all terms to the left:

42-((2x+1)(x+3))=0

We multiply parentheses ..

-((+2x^2+6x+x+3))+42=0


We calculate terms in parentheses: -((+2x^2+6x+x+3)), so:

(+2x^2+6x+x+3)

We get rid of parentheses

2x^2+6x+x+3

We add all the numbers together, and all the variables

2x^2+7x+3

Back to the equation:

-(2x^2+7x+3)


We get rid of parentheses

-2x^2-7x-3+42=0

We add all the numbers together, and all the variables

-2x^2-7x+39=0

a = -2; b = -7; c = +39;
Δ = b2-4ac
Δ = -72-4·(-2)·39
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-19}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+19}{2*-2}=\frac{26}{-4}
=-6+1/2
$