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2n^2-45n+252=0
a = 2; b = -45; c = +252;
Δ = b2-4ac
Δ = -452-4·2·252
Δ = 9
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{9}=3$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-3}{2*2}=\frac{42}{4} =10+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+3}{2*2}=\frac{48}{4} =12 $
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