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4p^2+3p-26=0
a = 4; b = 3; c = -26;
Δ = b2-4ac
Δ = 32-4·4·(-26)
Δ = 425
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5\sqrt{17}}{2*4}=\frac{-3-5\sqrt{17}}{8} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5\sqrt{17}}{2*4}=\frac{-3+5\sqrt{17}}{8} $
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