(r+9)(r-7)=0

Solution for (r+9)(r-7)=0 equation:

(r+9)(r-7)=0

We multiply parentheses ..

(+r^2-7r+9r-63)=0

We get rid of parentheses

r^2-7r+9r-63=0

We add all the numbers together, and all the variables

r^2+2r-63=0

a = 1; b = 2; c = -63;
Δ = b2-4ac
Δ = 22-4·1·(-63)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-16}{2*1}=\frac{-18}{2}
=-9
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+16}{2*1}=\frac{14}{2}
=7
$