165=(4x+9)(2x+8)

Solution for 165=(4x+9)(2x+8) equation:

165=(4x+9)(2x+8)

We move all terms to the left:

165-((4x+9)(2x+8))=0

We multiply parentheses ..

-((+8x^2+32x+18x+72))+165=0


We calculate terms in parentheses: -((+8x^2+32x+18x+72)), so:

(+8x^2+32x+18x+72)

We get rid of parentheses

8x^2+32x+18x+72

We add all the numbers together, and all the variables

8x^2+50x+72

Back to the equation:

-(8x^2+50x+72)


We get rid of parentheses

-8x^2-50x-72+165=0

We add all the numbers together, and all the variables

-8x^2-50x+93=0

a = -8; b = -50; c = +93;
Δ = b2-4ac
Δ = -502-4·(-8)·93
Δ = 5476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5476}=74$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-74}{2*-8}=\frac{-24}{-16}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+74}{2*-8}=\frac{124}{-16}
=-7+3/4
$