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5r^2+6r-11=0
a = 5; b = 6; c = -11;
Δ = b2-4ac
Δ = 62-4·5·(-11)
Δ = 256
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{256}=16$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-16}{2*5}=\frac{-22}{10} =-2+1/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+16}{2*5}=\frac{10}{10} =1 $
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