(4z-7)(2-z)=0

Solution for (4z-7)(2-z)=0 equation:

(4z-7)(2-z)=0

We add all the numbers together, and all the variables

(4z-7)(-1z+2)=0

We multiply parentheses ..

(-4z^2+8z+7z-14)=0

We get rid of parentheses

-4z^2+8z+7z-14=0

We add all the numbers together, and all the variables

-4z^2+15z-14=0

a = -4; b = 15; c = -14;
Δ = b2-4ac
Δ = 152-4·(-4)·(-14)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-1}{2*-4}=\frac{-16}{-8}
=+2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+1}{2*-4}=\frac{-14}{-8}
=1+3/4
$