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-3+9n^2=61
We move all terms to the left:
-3+9n^2-(61)=0
We add all the numbers together, and all the variables
9n^2-64=0
a = 9; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·9·(-64)
Δ = 2304
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{2304}=48$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*9}=\frac{-48}{18} =-2+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*9}=\frac{48}{18} =2+2/3 $
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