(6-z)(4z-7)=0

Solution for (6-z)(4z-7)=0 equation:

(6-z)(4z-7)=0

We add all the numbers together, and all the variables

(-1z+6)(4z-7)=0

We multiply parentheses ..

(-4z^2+7z+24z-42)=0

We get rid of parentheses

-4z^2+7z+24z-42=0

We add all the numbers together, and all the variables

-4z^2+31z-42=0

a = -4; b = 31; c = -42;
Δ = b2-4ac
Δ = 312-4·(-4)·(-42)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-17}{2*-4}=\frac{-48}{-8}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+17}{2*-4}=\frac{-14}{-8}
=1+3/4
$