(8r-5)(r-2)=0

Solution for (8r-5)(r-2)=0 equation:

(8r-5)(r-2)=0

We multiply parentheses ..

(+8r^2-16r-5r+10)=0

We get rid of parentheses

8r^2-16r-5r+10=0

We add all the numbers together, and all the variables

8r^2-21r+10=0

a = 8; b = -21; c = +10;
Δ = b2-4ac
Δ = -212-4·8·10
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{-(-21)-11}{2*8}=\frac{10}{16}
=5/8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+11}{2*8}=\frac{32}{16}
=2
$