(7+2x)(4+x)=55

Solution for (7+2x)(4+x)=55 equation:

(7+2x)(4+x)=55

We move all terms to the left:

(7+2x)(4+x)-(55)=0

We add all the numbers together, and all the variables

(2x+7)(x+4)-55=0

We multiply parentheses ..

(+2x^2+8x+7x+28)-55=0

We get rid of parentheses

2x^2+8x+7x+28-55=0

We add all the numbers together, and all the variables

2x^2+15x-27=0

a = 2; b = 15; c = -27;
Δ = b2-4ac
Δ = 152-4·2·(-27)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-21}{2*2}=\frac{-36}{4}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+21}{2*2}=\frac{6}{4}
=1+1/2
$