112=(x+1)(2x+4)

Solution for 112=(x+1)(2x+4) equation:

112=(x+1)(2x+4)

We move all terms to the left:

112-((x+1)(2x+4))=0

We multiply parentheses ..

-((+2x^2+4x+2x+4))+112=0


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

(+2x^2+4x+2x+4)

We get rid of parentheses

2x^2+4x+2x+4

We add all the numbers together, and all the variables

2x^2+6x+4

Back to the equation:

-(2x^2+6x+4)


We get rid of parentheses

-2x^2-6x-4+112=0

We add all the numbers together, and all the variables

-2x^2-6x+108=0

a = -2; b = -6; c = +108;
Δ = b2-4ac
Δ = -62-4·(-2)·108
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-30}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+30}{2*-2}=\frac{36}{-4}
=-9
$