(4z+3)(5z)=150

Solution for (4z+3)(5z)=150 equation:

(4z+3)(5z)=150

We move all terms to the left:

(4z+3)(5z)-(150)=0

We multiply parentheses

20z^2+15z-150=0

a = 20; b = 15; c = -150;
Δ = b2-4ac
Δ = 152-4·20·(-150)
Δ = 12225
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:

$\sqrt{\Delta}=\sqrt{12225}=\sqrt{25*489}=\sqrt{25}*\sqrt{489}=5\sqrt{489}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{489}}{2*20}=\frac{-15-5\sqrt{489}}{40}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{489}}{2*20}=\frac{-15+5\sqrt{489}}{40}
$