(7-z)(4z-9)=0

Solution for (7-z)(4z-9)=0 equation:

(7-z)(4z-9)=0

We add all the numbers together, and all the variables

(-1z+7)(4z-9)=0

We multiply parentheses ..

(-4z^2+9z+28z-63)=0

We get rid of parentheses

-4z^2+9z+28z-63=0

We add all the numbers together, and all the variables

-4z^2+37z-63=0

a = -4; b = 37; c = -63;
Δ = b2-4ac
Δ = 372-4·(-4)·(-63)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-19}{2*-4}=\frac{-56}{-8}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+19}{2*-4}=\frac{-18}{-8}
=2+1/4
$