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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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-32}{2*-5}=\frac{-60}{-10} =+6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+32}{2*-5}=\frac{4}{-10} =-2/5 $
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