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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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*5}=\frac{-40}{10} =-4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*5}=\frac{8}{10} =4/5 $
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