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4p^2+6p-48=0
a = 4; b = 6; c = -48;
Δ = b2-4ac
Δ = 62-4·4·(-48)
Δ = 804
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{201}}{2*4}=\frac{-6-2\sqrt{201}}{8} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{201}}{2*4}=\frac{-6+2\sqrt{201}}{8} $
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