21p+4p2=-20

Solution for 21p+4p2=-20 equation:

21p+4p^2=-20

We move all terms to the left:

21p+4p^2-(-20)=0

We add all the numbers together, and all the variables

4p^2+21p+20=0

a = 4; b = 21; c = +20;
Δ = b2-4ac
Δ = 212-4·4·20
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*4}=\frac{-32}{8}
=-4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*4}=\frac{-10}{8}
=-1+1/4
$