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3n^2=150
We move all terms to the left:
3n^2-(150)=0
a = 3; b = 0; c = -150;
Δ = b2-4ac
Δ = 02-4·3·(-150)
Δ = 1800
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{2}}{2*3}=\frac{0-30\sqrt{2}}{6} =-\frac{30\sqrt{2}}{6} =-5\sqrt{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{2}}{2*3}=\frac{0+30\sqrt{2}}{6} =\frac{30\sqrt{2}}{6} =5\sqrt{2} $
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