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9n^2+10=406
We move all terms to the left:
9n^2+10-(406)=0
We add all the numbers together, and all the variables
9n^2-396=0
a = 9; b = 0; c = -396;
Δ = b2-4ac
Δ = 02-4·9·(-396)
Δ = 14256
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14256}=\sqrt{1296*11}=\sqrt{1296}*\sqrt{11}=36\sqrt{11}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{11}}{2*9}=\frac{0-36\sqrt{11}}{18} =-\frac{36\sqrt{11}}{18} =-2\sqrt{11} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{11}}{2*9}=\frac{0+36\sqrt{11}}{18} =\frac{36\sqrt{11}}{18} =2\sqrt{11} $
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