(2z-7)(5-z)=0

Solution for (2z-7)(5-z)=0 equation:

(2z-7)(5-z)=0

We add all the numbers together, and all the variables

(2z-7)(-1z+5)=0

We multiply parentheses ..

(-2z^2+10z+7z-35)=0

We get rid of parentheses

-2z^2+10z+7z-35=0

We add all the numbers together, and all the variables

-2z^2+17z-35=0

a = -2; b = 17; c = -35;
Δ = b2-4ac
Δ = 172-4·(-2)·(-35)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-3}{2*-2}=\frac{-20}{-4}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+3}{2*-2}=\frac{-14}{-4}
=3+1/2
$