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20r^2+11r-3=0
a = 20; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·20·(-3)
Δ = 361
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{361}=19$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*20}=\frac{-30}{40} =-3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*20}=\frac{8}{40} =1/5 $
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