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5r^2+2r-16=0
a = 5; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·5·(-16)
Δ = 324
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{324}=18$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*5}=\frac{-20}{10} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*5}=\frac{16}{10} =1+3/5 $
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