4p^2+4p=99

Solution for 4p^2+4p=99 equation:

4p^2+4p=99

We move all terms to the left:

4p^2+4p-(99)=0

a = 4; b = 4; c = -99;
Δ = b2-4ac
Δ = 42-4·4·(-99)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-40}{2*4}=\frac{-44}{8}
=-5+1/2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+40}{2*4}=\frac{36}{8}
=4+1/2
$