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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4z^2+19z+63=0

a = -4; b = 19; c = +63;
Δ = b2-4ac
Δ = 192-4·(-4)·63
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-37}{2*-4}=\frac{-56}{-8}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+37}{2*-4}=\frac{18}{-8}
=-2+1/4
$