r^2-12r-64=0

Solution for r^2-12r-64=0 equation:

r^2-12r-64=0

a = 1; b = -12; c = -64;
Δ = b2-4ac
Δ = -122-4·1·(-64)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*1}=\frac{-8}{2}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*1}=\frac{32}{2}
=16
$