25x2+20x+4=5

Solution for 25x2+20x+4=5 equation:

25x^2+20x+4=5

We move all terms to the left:

25x^2+20x+4-(5)=0

We add all the numbers together, and all the variables

25x^2+20x-1=0

a = 25; b = 20; c = -1;
Δ = b2-4ac
Δ = 202-4·25·(-1)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{5}}{2*25}=\frac{-20-10\sqrt{5}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{5}}{2*25}=\frac{-20+10\sqrt{5}}{50}
$