(3r-2)(2r-5)+(r-7)(3r-2)=0

Solution for (3r-2)(2r-5)+(r-7)(3r-2)=0 equation:

(3r-2)(2r-5)+(r-7)(3r-2)=0

We multiply parentheses ..

(+6r^2-15r-4r+10)+(r-7)(3r-2)=0

We get rid of parentheses

6r^2-15r-4r+(r-7)(3r-2)+10=0

We multiply parentheses ..

6r^2+(+3r^2-2r-21r+14)-15r-4r+10=0

We add all the numbers together, and all the variables

6r^2+(+3r^2-2r-21r+14)-19r+10=0

We get rid of parentheses

6r^2+3r^2-2r-21r-19r+14+10=0

We add all the numbers together, and all the variables

9r^2-42r+24=0

a = 9; b = -42; c = +24;
Δ = b2-4ac
Δ = -422-4·9·24
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-30}{2*9}=\frac{12}{18}
=2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+30}{2*9}=\frac{72}{18}
=4
$