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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4z^2+5z+12z+15)=0

We get rid of parentheses

4z^2+5z+12z+15=0

We add all the numbers together, and all the variables

4z^2+17z+15=0

a = 4; b = 17; c = +15;
Δ = b2-4ac
Δ = 172-4·4·15
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*4}=\frac{-24}{8}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*4}=\frac{-10}{8}
=-1+1/4
$