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5p^2+28p+32=0
a = 5; b = 28; c = +32;
Δ = b2-4ac
Δ = 282-4·5·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12}{2*5}=\frac{-40}{10} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12}{2*5}=\frac{-16}{10} =-1+3/5 $
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