(4z+3)(5z)=96

Solution for (4z+3)(5z)=96 equation:

(4z+3)(5z)=96

We move all terms to the left:

(4z+3)(5z)-(96)=0

We multiply parentheses

20z^2+15z-96=0

a = 20; b = 15; c = -96;
Δ = b2-4ac
Δ = 152-4·20·(-96)
Δ = 7905
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{-(15)-\sqrt{7905}}{2*20}=\frac{-15-\sqrt{7905}}{40}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{7905}}{2*20}=\frac{-15+\sqrt{7905}}{40}
$