-18=(r-4)(r+5)

Solution for -18=(r-4)(r+5) equation:

-18=(r-4)(r+5)

We move all terms to the left:

-18-((r-4)(r+5))=0

We multiply parentheses ..

-((+r^2+5r-4r-20))-18=0


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

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

We get rid of parentheses

r^2+5r-4r-20

We add all the numbers together, and all the variables

r^2+r-20

Back to the equation:

-(r^2+r-20)


We get rid of parentheses

-r^2-r+20-18=0

We add all the numbers together, and all the variables

-1r^2-1r+2=0

a = -1; b = -1; c = +2;
Δ = b2-4ac
Δ = -12-4·(-1)·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*-1}=\frac{-2}{-2}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*-1}=\frac{4}{-2}
=-2
$