10p^2+21p-108=0

Solution for 10p^2+21p-108=0 equation:

10p^2+21p-108=0

a = 10; b = 21; c = -108;
Δ = b2-4ac
Δ = 212-4·10·(-108)
Δ = 4761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4761}=69$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-69}{2*10}=\frac{-90}{20}
=-4+1/2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+69}{2*10}=\frac{48}{20}
=2+2/5
$