8(9+r^2)-64=3r(3r+1)-r

Solution for 8(9+r^2)-64=3r(3r+1)-r equation:

8(9+r^2)-64=3r(3r+1)-r

We move all terms to the left:

8(9+r^2)-64-(3r(3r+1)-r)=0

We multiply parentheses

8r^2-(3r(3r+1)-r)+72-64=0


We calculate terms in parentheses: -(3r(3r+1)-r), so:

3r(3r+1)-r

We add all the numbers together, and all the variables

-1r+3r(3r+1)

We multiply parentheses

9r^2-1r+3r

We add all the numbers together, and all the variables

9r^2+2r

Back to the equation:

-(9r^2+2r)


We add all the numbers together, and all the variables

8r^2-(9r^2+2r)+8=0

We get rid of parentheses

8r^2-9r^2-2r+8=0

We add all the numbers together, and all the variables

-1r^2-2r+8=0

a = -1; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-1}=\frac{-4}{-2}
=+2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-1}=\frac{8}{-2}
=-4
$