4r^2+r=13r

Solution for 4r^2+r=13r equation:

4r^2+r=13r

We move all terms to the left:

4r^2+r-(13r)=0

We add all the numbers together, and all the variables

4r^2-12r=0

a = 4; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·4·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*4}=\frac{0}{8}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*4}=\frac{24}{8}
=3
$