(3z+1)(4z-1)=70

Solution for (3z+1)(4z-1)=70 equation:

(3z+1)(4z-1)=70

We move all terms to the left:

(3z+1)(4z-1)-(70)=0

We multiply parentheses ..

(+12z^2-3z+4z-1)-70=0

We get rid of parentheses

12z^2-3z+4z-1-70=0

We add all the numbers together, and all the variables

12z^2+z-71=0

a = 12; b = 1; c = -71;
Δ = b2-4ac
Δ = 12-4·12·(-71)
Δ = 3409
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{-(1)-\sqrt{3409}}{2*12}=\frac{-1-\sqrt{3409}}{24}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3409}}{2*12}=\frac{-1+\sqrt{3409}}{24}
$