25x^2-20x+4=27

Solution for 25x^2-20x+4=27 equation:

25x^2-20x+4=27

We move all terms to the left:

25x^2-20x+4-(27)=0

We add all the numbers together, and all the variables

25x^2-20x-23=0

a = 25; b = -20; c = -23;
Δ = b2-4ac
Δ = -202-4·25·(-23)
Δ = 2700
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:

$\sqrt{\Delta}=\sqrt{2700}=\sqrt{900*3}=\sqrt{900}*\sqrt{3}=30\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-30\sqrt{3}}{2*25}=\frac{20-30\sqrt{3}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+30\sqrt{3}}{2*25}=\frac{20+30\sqrt{3}}{50}
$