2p^2+7p-15=0

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Solution for 2p^2+7p-15=0 equation:



2p^2+7p-15=0
a = 2; b = 7; c = -15;
Δ = b2-4ac
Δ = 72-4·2·(-15)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*2}=\frac{-20}{4} =-5 $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*2}=\frac{6}{4} =1+1/2 $

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