-2p2-15p+25=0

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Solution for -2p2-15p+25=0 equation:



-2p^2-15p+25=0
a = -2; b = -15; c = +25;
Δ = b2-4ac
Δ = -152-4·(-2)·25
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{-(-15)-5\sqrt{17}}{2*-2}=\frac{15-5\sqrt{17}}{-4} $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{17}}{2*-2}=\frac{15+5\sqrt{17}}{-4} $

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