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

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

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

We add all the numbers together, and all the variables

(-1z+6)(4z+5)=0

We multiply parentheses ..

(-4z^2-5z+24z+30)=0

We get rid of parentheses

-4z^2-5z+24z+30=0

We add all the numbers together, and all the variables

-4z^2+19z+30=0

a = -4; b = 19; c = +30;
Δ = b2-4ac
Δ = 192-4·(-4)·30
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-29}{2*-4}=\frac{-48}{-8}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+29}{2*-4}=\frac{10}{-8}
=-1+1/4
$