(2r-5)(10r-7)=0

Solution for (2r-5)(10r-7)=0 equation:

(2r-5)(10r-7)=0

We multiply parentheses ..

(+20r^2-14r-50r+35)=0

We get rid of parentheses

20r^2-14r-50r+35=0

We add all the numbers together, and all the variables

20r^2-64r+35=0

a = 20; b = -64; c = +35;
Δ = b2-4ac
Δ = -642-4·20·35
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-36}{2*20}=\frac{28}{40}
=7/10
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+36}{2*20}=\frac{100}{40}
=2+1/2
$