25p^2+40p+16=12

Solution for 25p^2+40p+16=12 equation:

25p^2+40p+16=12

We move all terms to the left:

25p^2+40p+16-(12)=0

We add all the numbers together, and all the variables

25p^2+40p+4=0

a = 25; b = 40; c = +4;
Δ = b2-4ac
Δ = 402-4·25·4
Δ = 1200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{3}}{2*25}=\frac{-40-20\sqrt{3}}{50}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{3}}{2*25}=\frac{-40+20\sqrt{3}}{50}
$