63=z(z-3)

Solution for 63=z(z-3) equation:

63=z(z-3)

We move all terms to the left:

63-(z(z-3))=0


We calculate terms in parentheses: -(z(z-3)), so:

z(z-3)

We multiply parentheses

z^2-3z

Back to the equation:

-(z^2-3z)


We get rid of parentheses

-z^2+3z+63=0

We add all the numbers together, and all the variables

-1z^2+3z+63=0

a = -1; b = 3; c = +63;
Δ = b2-4ac
Δ = 32-4·(-1)·63
Δ = 261
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{29}}{2*-1}=\frac{-3-3\sqrt{29}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{29}}{2*-1}=\frac{-3+3\sqrt{29}}{-2}
$