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3n^2-300=0
a = 3; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·3·(-300)
Δ = 3600
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*3}=\frac{-60}{6} =-10 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*3}=\frac{60}{6} =10 $
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