15p^2+100p-160=0

Solution for 15p^2+100p-160=0 equation:

15p^2+100p-160=0

a = 15; b = 100; c = -160;
Δ = b2-4ac
Δ = 1002-4·15·(-160)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-140}{2*15}=\frac{-240}{30}
=-8
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+140}{2*15}=\frac{40}{30}
=1+1/3
$