165=(3x+5)*(4x-7)

Solution for 165=(3x+5)*(4x-7) equation:

165=(3x+5)(4x-7)

We move all terms to the left:

165-((3x+5)(4x-7))=0

We multiply parentheses ..

-((+12x^2-21x+20x-35))+165=0


We calculate terms in parentheses: -((+12x^2-21x+20x-35)), so:

(+12x^2-21x+20x-35)

We get rid of parentheses

12x^2-21x+20x-35

We add all the numbers together, and all the variables

12x^2-1x-35

Back to the equation:

-(12x^2-1x-35)


We get rid of parentheses

-12x^2+1x+35+165=0

We add all the numbers together, and all the variables

-12x^2+x+200=0

a = -12; b = 1; c = +200;
Δ = b2-4ac
Δ = 12-4·(-12)·200
Δ = 9601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{9601}}{2*-12}=\frac{-1-\sqrt{9601}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{9601}}{2*-12}=\frac{-1+\sqrt{9601}}{-24}
$