63=(4-x)(5-x)

Solution for 63=(4-x)(5-x) equation:

63=(4-x)(5-x)

We move all terms to the left:

63-((4-x)(5-x))=0

We add all the numbers together, and all the variables

-((-1x+4)(-1x+5))+63=0

We multiply parentheses ..

-((+x^2-5x-4x+20))+63=0


We calculate terms in parentheses: -((+x^2-5x-4x+20)), so:

(+x^2-5x-4x+20)

We get rid of parentheses

x^2-5x-4x+20

We add all the numbers together, and all the variables

x^2-9x+20

Back to the equation:

-(x^2-9x+20)


We get rid of parentheses

-x^2+9x-20+63=0

We add all the numbers together, and all the variables

-1x^2+9x+43=0

a = -1; b = 9; c = +43;
Δ = b2-4ac
Δ = 92-4·(-1)·43
Δ = 253
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{253}}{2*-1}=\frac{-9-\sqrt{253}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{253}}{2*-1}=\frac{-9+\sqrt{253}}{-2}
$