25a^2+20a+4=

Solution for 25a^2+20a+4= equation:

25a^2+20a+4=

We move all terms to the left:

25a^2+20a+4-()=0

We add all the numbers together, and all the variables

25a^2+20a=0

a = 25; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·25·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*25}=\frac{-40}{50}
=-4/5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*25}=\frac{0}{50}
=0
$