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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

12x^2+16x+5=0

a = 12; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·12·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*12}=\frac{-20}{24}
=-5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*12}=\frac{-12}{24}
=-1/2
$