21p^2+18p-28p-24=0

Solution for 21p^2+18p-28p-24=0 equation:

21p^2+18p-28p-24=0

We add all the numbers together, and all the variables

21p^2-10p-24=0

a = 21; b = -10; c = -24;
Δ = b2-4ac
Δ = -102-4·21·(-24)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-46}{2*21}=\frac{-36}{42}
=-6/7
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+46}{2*21}=\frac{56}{42}
=1+1/3
$