56=(x+3)(x+4)

Solution for 56=(x+3)(x+4) equation:

56=(x+3)(x+4)

We move all terms to the left:

56-((x+3)(x+4))=0

We multiply parentheses ..

-((+x^2+4x+3x+12))+56=0


We calculate terms in parentheses: -((+x^2+4x+3x+12)), so:

(+x^2+4x+3x+12)

We get rid of parentheses

x^2+4x+3x+12

We add all the numbers together, and all the variables

x^2+7x+12

Back to the equation:

-(x^2+7x+12)


We get rid of parentheses

-x^2-7x-12+56=0

We add all the numbers together, and all the variables

-1x^2-7x+44=0

a = -1; b = -7; c = +44;
Δ = b2-4ac
Δ = -72-4·(-1)·44
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-15}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+15}{2*-1}=\frac{22}{-2}
=-11
$