(4z+3)(6z+5)=0

Solution for (4z+3)(6z+5)=0 equation:

(4z+3)(6z+5)=0

We multiply parentheses ..

(+24z^2+20z+18z+15)=0

We get rid of parentheses

24z^2+20z+18z+15=0

We add all the numbers together, and all the variables

24z^2+38z+15=0

a = 24; b = 38; c = +15;
Δ = b2-4ac
Δ = 382-4·24·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2}{2*24}=\frac{-40}{48}
=-5/6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2}{2*24}=\frac{-36}{48}
=-3/4
$