(3x-5)(4x+1)=77

Solution for (3x-5)(4x+1)=77 equation:

(3x-5)(4x+1)=77

We move all terms to the left:

(3x-5)(4x+1)-(77)=0

We multiply parentheses ..

(+12x^2+3x-20x-5)-77=0

We get rid of parentheses

12x^2+3x-20x-5-77=0

We add all the numbers together, and all the variables

12x^2-17x-82=0

a = 12; b = -17; c = -82;
Δ = b2-4ac
Δ = -172-4·12·(-82)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-65}{2*12}=\frac{-48}{24}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+65}{2*12}=\frac{82}{24}
=3+5/12
$