25x^2+20x+4=50

Solution for 25x^2+20x+4=50 equation:

25x^2+20x+4=50

We move all terms to the left:

25x^2+20x+4-(50)=0

We add all the numbers together, and all the variables

25x^2+20x-46=0

a = 25; b = 20; c = -46;
Δ = b2-4ac
Δ = 202-4·25·(-46)
Δ = 5000
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:

$\sqrt{\Delta}=\sqrt{5000}=\sqrt{2500*2}=\sqrt{2500}*\sqrt{2}=50\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-50\sqrt{2}}{2*25}=\frac{-20-50\sqrt{2}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+50\sqrt{2}}{2*25}=\frac{-20+50\sqrt{2}}{50}
$