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4n^2-4n-63=0
a = 4; b = -4; c = -63;
Δ = b2-4ac
Δ = -42-4·4·(-63)
Δ = 1024
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{1024}=32$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-32}{2*4}=\frac{-28}{8} =-3+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+32}{2*4}=\frac{36}{8} =4+1/2 $
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