r=2r^2-12r+16

Solution for r=2r^2-12r+16 equation:

r=2r^2-12r+16

We move all terms to the left:

r-(2r^2-12r+16)=0

We get rid of parentheses

-2r^2+r+12r-16=0

We add all the numbers together, and all the variables

-2r^2+13r-16=0

a = -2; b = 13; c = -16;
Δ = b2-4ac
Δ = 132-4·(-2)·(-16)
Δ = 41
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{-(13)-\sqrt{41}}{2*-2}=\frac{-13-\sqrt{41}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{41}}{2*-2}=\frac{-13+\sqrt{41}}{-4}
$