(7-z)(2z+1)=0

Solution for (7-z)(2z+1)=0 equation:

(7-z)(2z+1)=0

We add all the numbers together, and all the variables

(-1z+7)(2z+1)=0

We multiply parentheses ..

(-2z^2-1z+14z+7)=0

We get rid of parentheses

-2z^2-1z+14z+7=0

We add all the numbers together, and all the variables

-2z^2+13z+7=0

a = -2; b = 13; c = +7;
Δ = b2-4ac
Δ = 132-4·(-2)·7
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-15}{2*-2}=\frac{-28}{-4}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+15}{2*-2}=\frac{2}{-4}
=-1/2
$