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2n^2+4n-14=286
We move all terms to the left:
2n^2+4n-14-(286)=0
We add all the numbers together, and all the variables
2n^2+4n-300=0
a = 2; b = 4; c = -300;
Δ = b2-4ac
Δ = 42-4·2·(-300)
Δ = 2416
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{151}}{2*2}=\frac{-4-4\sqrt{151}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{151}}{2*2}=\frac{-4+4\sqrt{151}}{4} $
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