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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+12x^2-3x+4x-1)=0

We get rid of parentheses

12x^2-3x+4x-1=0

We add all the numbers together, and all the variables

12x^2+x-1=0

a = 12; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·12·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*12}=\frac{-8}{24}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*12}=\frac{6}{24}
=1/4
$