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

Solution for (3+z)(4z-1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4z^2+11z-3=0

a = 4; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·4·(-3)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*4}=\frac{-24}{8}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*4}=\frac{2}{8}
=1/4
$