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4p^2+19p-3=0
a = 4; b = 19; c = -3;
Δ = b2-4ac
Δ = 192-4·4·(-3)
Δ = 409
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{409}}{2*4}=\frac{-19-\sqrt{409}}{8} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{409}}{2*4}=\frac{-19+\sqrt{409}}{8} $
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