(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

(z+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
$