25x^2+200x+4=0

Solution for 25x^2+200x+4=0 equation:

25x^2+200x+4=0

a = 25; b = 200; c = +4;
Δ = b2-4ac
Δ = 2002-4·25·4
Δ = 39600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{39600}=\sqrt{3600*11}=\sqrt{3600}*\sqrt{11}=60\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-60\sqrt{11}}{2*25}=\frac{-200-60\sqrt{11}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+60\sqrt{11}}{2*25}=\frac{-200+60\sqrt{11}}{50}
$