2p^2+p=1

Solution for 2p^2+p=1 equation:

2p^2+p=1

We move all terms to the left:

2p^2+p-(1)=0

a = 2; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·2·(-1)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*2}=\frac{-4}{4}
=-1
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*2}=\frac{2}{4}
=1/2
$