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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4z^2+28z-7z-49)=0

We get rid of parentheses

4z^2+28z-7z-49=0

We add all the numbers together, and all the variables

4z^2+21z-49=0

a = 4; b = 21; c = -49;
Δ = b2-4ac
Δ = 212-4·4·(-49)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-35}{2*4}=\frac{-56}{8}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+35}{2*4}=\frac{14}{8}
=1+3/4
$