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35n^2-22n+3=0
a = 35; b = -22; c = +3;
Δ = b2-4ac
Δ = -222-4·35·3
Δ = 64
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{64}=8$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-8}{2*35}=\frac{14}{70} =1/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+8}{2*35}=\frac{30}{70} =3/7 $
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