4z(5z+4)=15z+12

Solution for 4z(5z+4)=15z+12 equation:

4z(5z+4)=15z+12

We move all terms to the left:

4z(5z+4)-(15z+12)=0

We multiply parentheses

20z^2+16z-(15z+12)=0

We get rid of parentheses

20z^2+16z-15z-12=0

We add all the numbers together, and all the variables

20z^2+z-12=0

a = 20; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·20·(-12)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*20}=\frac{-32}{40}
=-4/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*20}=\frac{30}{40}
=3/4
$