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5r^2-16r-16=0
a = 5; b = -16; c = -16;
Δ = b2-4ac
Δ = -162-4·5·(-16)
Δ = 576
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{576}=24$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24}{2*5}=\frac{-8}{10} =-4/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24}{2*5}=\frac{40}{10} =4 $
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