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4p^2-16p-9=0
a = 4; b = -16; c = -9;
Δ = b2-4ac
Δ = -162-4·4·(-9)
Δ = 400
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{400}=20$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*4}=\frac{-4}{8} =-1/2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*4}=\frac{36}{8} =4+1/2 $
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