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60p^2+4p-160=0
a = 60; b = 4; c = -160;
Δ = b2-4ac
Δ = 42-4·60·(-160)
Δ = 38416
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{38416}=196$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-196}{2*60}=\frac{-200}{120} =-1+2/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+196}{2*60}=\frac{192}{120} =1+3/5 $
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