(2x+1)(6x+5)=1

Solution for (2x+1)(6x+5)=1 equation:

(2x+1)(6x+5)=1

We move all terms to the left:

(2x+1)(6x+5)-(1)=0

We multiply parentheses ..

(+12x^2+10x+6x+5)-1=0

We get rid of parentheses

12x^2+10x+6x+5-1=0

We add all the numbers together, and all the variables

12x^2+16x+4=0

a = 12; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·12·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*12}=\frac{-24}{24}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*12}=\frac{-8}{24}
=-1/3
$