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2n^2-10n-300=0
a = 2; b = -10; c = -300;
Δ = b2-4ac
Δ = -102-4·2·(-300)
Δ = 2500
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{2500}=50$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-50}{2*2}=\frac{-40}{4} =-10 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+50}{2*2}=\frac{60}{4} =15 $
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