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2n^2+20n=30
We move all terms to the left:
2n^2+20n-(30)=0
a = 2; b = 20; c = -30;
Δ = b2-4ac
Δ = 202-4·2·(-30)
Δ = 640
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{10}}{2*2}=\frac{-20-8\sqrt{10}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{10}}{2*2}=\frac{-20+8\sqrt{10}}{4} $
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