92=(3z+3)(4z+1)

Solution for 92=(3z+3)(4z+1) equation:

92=(3z+3)(4z+1)

We move all terms to the left:

92-((3z+3)(4z+1))=0

We multiply parentheses ..

-((+12z^2+3z+12z+3))+92=0


We calculate terms in parentheses: -((+12z^2+3z+12z+3)), so:

(+12z^2+3z+12z+3)

We get rid of parentheses

12z^2+3z+12z+3

We add all the numbers together, and all the variables

12z^2+15z+3

Back to the equation:

-(12z^2+15z+3)


We get rid of parentheses

-12z^2-15z-3+92=0

We add all the numbers together, and all the variables

-12z^2-15z+89=0

a = -12; b = -15; c = +89;
Δ = b2-4ac
Δ = -152-4·(-12)·89
Δ = 4497
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{4497}}{2*-12}=\frac{15-\sqrt{4497}}{-24}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{4497}}{2*-12}=\frac{15+\sqrt{4497}}{-24}
$