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2n^2+3n-504=0
a = 2; b = 3; c = -504;
Δ = b2-4ac
Δ = 32-4·2·(-504)
Δ = 4041
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4041}=\sqrt{9*449}=\sqrt{9}*\sqrt{449}=3\sqrt{449}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{449}}{2*2}=\frac{-3-3\sqrt{449}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{449}}{2*2}=\frac{-3+3\sqrt{449}}{4} $
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