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5r^2+3r-36=0
a = 5; b = 3; c = -36;
Δ = b2-4ac
Δ = 32-4·5·(-36)
Δ = 729
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{729}=27$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*5}=\frac{-30}{10} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*5}=\frac{24}{10} =2+2/5 $
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