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20x^2-11x-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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*20}=\frac{-8}{40} =-1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*20}=\frac{30}{40} =3/4 $
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