28r=r(r+3)

Solution for 28r=r(r+3) equation:

28r=r(r+3)

We move all terms to the left:

28r-(r(r+3))=0


We calculate terms in parentheses: -(r(r+3)), so:

r(r+3)

We multiply parentheses

r^2+3r

Back to the equation:

-(r^2+3r)


We get rid of parentheses

-r^2+28r-3r=0

We add all the numbers together, and all the variables

-1r^2+25r=0

a = -1; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-1)·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-1}=\frac{-50}{-2}
=+25
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-1}=\frac{0}{-2}
=0
$