(4x+1)(3x+1)=13

Solution for (4x+1)(3x+1)=13 equation:

(4x+1)(3x+1)=13

We move all terms to the left:

(4x+1)(3x+1)-(13)=0

We multiply parentheses ..

(+12x^2+4x+3x+1)-13=0

We get rid of parentheses

12x^2+4x+3x+1-13=0

We add all the numbers together, and all the variables

12x^2+7x-12=0

a = 12; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·12·(-12)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-25}{2*12}=\frac{-32}{24}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+25}{2*12}=\frac{18}{24}
=3/4
$