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7p^2+27p-40=0
a = 7; b = 27; c = -40;
Δ = b2-4ac
Δ = 272-4·7·(-40)
Δ = 1849
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{1849}=43$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-43}{2*7}=\frac{-70}{14} =-5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+43}{2*7}=\frac{16}{14} =1+1/7 $
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