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4n^2-14n+10=0
a = 4; b = -14; c = +10;
Δ = b2-4ac
Δ = -142-4·4·10
Δ = 36
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{36}=6$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*4}=\frac{8}{8} =1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*4}=\frac{20}{8} =2+1/2 $
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