16-(2z+1)z=-4

Solution for 16-(2z+1)z=-4 equation:

16-(2z+1)z=-4

We move all terms to the left:

16-(2z+1)z-(-4)=0

We add all the numbers together, and all the variables

-(2z+1)z+20=0

We multiply parentheses

-2z^2-z+20=0

We add all the numbers together, and all the variables

-2z^2-1z+20=0

a = -2; b = -1; c = +20;
Δ = b2-4ac
Δ = -12-4·(-2)·20
Δ = 161
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{161}}{2*-2}=\frac{1-\sqrt{161}}{-4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{161}}{2*-2}=\frac{1+\sqrt{161}}{-4}
$