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5n^2-28n-12=0
a = 5; b = -28; c = -12;
Δ = b2-4ac
Δ = -282-4·5·(-12)
Δ = 1024
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{1024}=32$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-32}{2*5}=\frac{-4}{10} =-2/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+32}{2*5}=\frac{60}{10} =6 $
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