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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4z^2+16z-3z-12)=0

We get rid of parentheses

4z^2+16z-3z-12=0

We add all the numbers together, and all the variables

4z^2+13z-12=0

a = 4; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·4·(-12)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*4}=\frac{-32}{8}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*4}=\frac{6}{8}
=3/4
$