(4z-5)(4+2z)=0

Solution for (4z-5)(4+2z)=0 equation:

(4z-5)(4+2z)=0

We add all the numbers together, and all the variables

(4z-5)(2z+4)=0

We multiply parentheses ..

(+8z^2+16z-10z-20)=0

We get rid of parentheses

8z^2+16z-10z-20=0

We add all the numbers together, and all the variables

8z^2+6z-20=0

a = 8; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·8·(-20)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-26}{2*8}=\frac{-32}{16}
=-2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+26}{2*8}=\frac{20}{16}
=1+1/4
$