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5p^2-18p+16=0
a = 5; b = -18; c = +16;
Δ = b2-4ac
Δ = -182-4·5·16
Δ = 4
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{4}=2$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*5}=\frac{16}{10} =1+3/5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*5}=\frac{20}{10} =2 $
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