(6+z)(4z+9)=0

Solution for (6+z)(4z+9)=0 equation:

(6+z)(4z+9)=0

We add all the numbers together, and all the variables

(z+6)(4z+9)=0

We multiply parentheses ..

(+4z^2+9z+24z+54)=0

We get rid of parentheses

4z^2+9z+24z+54=0

We add all the numbers together, and all the variables

4z^2+33z+54=0

a = 4; b = 33; c = +54;
Δ = b2-4ac
Δ = 332-4·4·54
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-15}{2*4}=\frac{-48}{8}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+15}{2*4}=\frac{-18}{8}
=-2+1/4
$