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-4p^2+28p-48=0
a = -4; b = 28; c = -48;
Δ = b2-4ac
Δ = 282-4·(-4)·(-48)
Δ = 16
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{16}=4$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*-4}=\frac{-32}{-8} =+4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*-4}=\frac{-24}{-8} =+3 $
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