(Z-4)(z-7)=z2-11z+28

Solution for (Z-4)(z-7)=z2-11z+28 equation:

(-4)(Z-7)=Z2-11Z+28

We move all terms to the left:

(-4)(Z-7)-(Z2-11Z+28)=0

We add all the numbers together, and all the variables

-(+Z^2-11Z+28)+(-4)(Z-7)=0

We get rid of parentheses

-Z^2+11Z+(-4)(Z-7)-28=0

We multiply parentheses ..

-Z^2+11Z+(-4Z+28)-28=0

We add all the numbers together, and all the variables

-1Z^2+11Z+(-4Z+28)-28=0

We get rid of parentheses

-1Z^2+11Z-4Z+28-28=0

We add all the numbers together, and all the variables

-1Z^2+7Z=0

a = -1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-1)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-1}=\frac{-14}{-2}
=+7
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-1}=\frac{0}{-2}
=0
$