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5n^2-4n-33=0
a = 5; b = -4; c = -33;
Δ = b2-4ac
Δ = -42-4·5·(-33)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-26}{2*5}=\frac{-22}{10} =-2+1/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+26}{2*5}=\frac{30}{10} =3 $
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