(r-3)2=(r2-5r)+3

Solution for (r-3)2=(r2-5r)+3 equation:

(r-3)2=(r2-5r)+3

We move all terms to the left:

(r-3)2-((r2-5r)+3)=0

We add all the numbers together, and all the variables

-((+r^2-5r)+3)+(r-3)2=0

We multiply parentheses

-((+r^2-5r)+3)+2r-6=0


We calculate terms in parentheses: -((+r^2-5r)+3), so:

(+r^2-5r)+3

We get rid of parentheses

r^2-5r+3

Back to the equation:

-(r^2-5r+3)


We add all the numbers together, and all the variables

2r-(r^2-5r+3)-6=0

We get rid of parentheses

-r^2+2r+5r-3-6=0

We add all the numbers together, and all the variables

-1r^2+7r-9=0

a = -1; b = 7; c = -9;
Δ = b2-4ac
Δ = 72-4·(-1)·(-9)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{13}}{2*-1}=\frac{-7-\sqrt{13}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{13}}{2*-1}=\frac{-7+\sqrt{13}}{-2}
$