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4n^2+32n+28=0
a = 4; b = 32; c = +28;
Δ = b2-4ac
Δ = 322-4·4·28
Δ = 576
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{576}=24$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-24}{2*4}=\frac{-56}{8} =-7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+24}{2*4}=\frac{-8}{8} =-1 $
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