p^2+1.2p-0.36=0

Solution for p^2+1.2p-0.36=0 equation:

p^2+1.2p-0.36=0

a = 1; b = 1.2; c = -0.36;
Δ = b2-4ac
Δ = 1.22-4·1·(-0.36)
Δ = 2.88
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{-(1.2)-\sqrt{2.88}}{2*1}=\frac{-1.2-\sqrt{2.88}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{2.88}}{2*1}=\frac{-1.2+\sqrt{2.88}}{2}
$