2p^2-3p-20=0

Solution for 2p^2-3p-20=0 equation:

2p^2-3p-20=0

a = 2; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·2·(-20)
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{-(-3)-13}{2*2}=\frac{-10}{4}
=-2+1/2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*2}=\frac{16}{4}
=4
$