r^2-4r-54=-8

Solution for r^2-4r-54=-8 equation:

r^2-4r-54=-8

We move all terms to the left:

r^2-4r-54-(-8)=0

We add all the numbers together, and all the variables

r^2-4r-46=0

a = 1; b = -4; c = -46;
Δ = b2-4ac
Δ = -42-4·1·(-46)
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10\sqrt{2}}{2*1}=\frac{4-10\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10\sqrt{2}}{2*1}=\frac{4+10\sqrt{2}}{2}
$