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5r^2+4r-9=0
a = 5; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·5·(-9)
Δ = 196
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{196}=14$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*5}=\frac{-18}{10} =-1+4/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*5}=\frac{10}{10} =1 $
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