r2-4r=0

Solution for r2-4r=0 equation:

r2-4r=0

We add all the numbers together, and all the variables

r^2-4r=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*1}=\frac{0}{2}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$