4z(z+5)=32

Solution for 4z(z+5)=32 equation:

4z(z+5)=32

We move all terms to the left:

4z(z+5)-(32)=0

We multiply parentheses

4z^2+20z-32=0

a = 4; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·4·(-32)
Δ = 912
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{57}}{2*4}=\frac{-20-4\sqrt{57}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{57}}{2*4}=\frac{-20+4\sqrt{57}}{8}
$