2r(r-10)=5

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

2r(r-10)=5

We move all terms to the left:

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

We multiply parentheses

2r^2-20r-5=0

a = 2; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·2·(-5)
Δ = 440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{110}}{2*2}=\frac{20-2\sqrt{110}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{110}}{2*2}=\frac{20+2\sqrt{110}}{4}
$