p^2+p=1.6

Solution for p^2+p=1.6 equation:

p^2+p=1.6

We move all terms to the left:

p^2+p-(1.6)=0

We add all the numbers together, and all the variables

p^2+p-1.6=0

a = 1; b = 1; c = -1.6;
Δ = b2-4ac
Δ = 12-4·1·(-1.6)
Δ = 7.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{7.4}}{2*1}=\frac{-1-\sqrt{7.4}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{7.4}}{2*1}=\frac{-1+\sqrt{7.4}}{2}
$