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10n^2+2=292
We move all terms to the left:
10n^2+2-(292)=0
We add all the numbers together, and all the variables
10n^2-290=0
a = 10; b = 0; c = -290;
Δ = b2-4ac
Δ = 02-4·10·(-290)
Δ = 11600
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11600}=\sqrt{400*29}=\sqrt{400}*\sqrt{29}=20\sqrt{29}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{29}}{2*10}=\frac{0-20\sqrt{29}}{20} =-\frac{20\sqrt{29}}{20} =-\sqrt{29} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{29}}{2*10}=\frac{0+20\sqrt{29}}{20} =\frac{20\sqrt{29}}{20} =\sqrt{29} $
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