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4n^2=36
We move all terms to the left:
4n^2-(36)=0
a = 4; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·4·(-36)
Δ = 576
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{576}=24$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*4}=\frac{-24}{8} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*4}=\frac{24}{8} =3 $
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