(r+5)(r-2)=18

Solution for (r+5)(r-2)=18 equation:

(r+5)(r-2)=18

We move all terms to the left:

(r+5)(r-2)-(18)=0

We multiply parentheses ..

(+r^2-2r+5r-10)-18=0

We get rid of parentheses

r^2-2r+5r-10-18=0

We add all the numbers together, and all the variables

r^2+3r-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$