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3n^2-12n-10=0
a = 3; b = -12; c = -10;
Δ = b2-4ac
Δ = -122-4·3·(-10)
Δ = 264
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{66}}{2*3}=\frac{12-2\sqrt{66}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{66}}{2*3}=\frac{12+2\sqrt{66}}{6} $
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