(r-5)(r+3)+(r-5)(r+2)=0

Solution for (r-5)(r+3)+(r-5)(r+2)=0 equation:

(r-5)(r+3)+(r-5)(r+2)=0

We multiply parentheses ..

(+r^2+3r-5r-15)+(r-5)(r+2)=0

We get rid of parentheses

r^2+3r-5r+(r-5)(r+2)-15=0

We multiply parentheses ..

r^2+(+r^2+2r-5r-10)+3r-5r-15=0

We add all the numbers together, and all the variables

r^2+(+r^2+2r-5r-10)-2r-15=0

We get rid of parentheses

r^2+r^2+2r-5r-2r-10-15=0

We add all the numbers together, and all the variables

2r^2-5r-25=0

a = 2; b = -5; c = -25;
Δ = b2-4ac
Δ = -52-4·2·(-25)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-15}{2*2}=\frac{-10}{4}
=-2+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+15}{2*2}=\frac{20}{4}
=5
$