25x^2-100x+49=0

Solution for 25x^2-100x+49=0 equation:

25x^2-100x+49=0

a = 25; b = -100; c = +49;
Δ = b2-4ac
Δ = -1002-4·25·49
Δ = 5100
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:

$\sqrt{\Delta}=\sqrt{5100}=\sqrt{100*51}=\sqrt{100}*\sqrt{51}=10\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-10\sqrt{51}}{2*25}=\frac{100-10\sqrt{51}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+10\sqrt{51}}{2*25}=\frac{100+10\sqrt{51}}{50}
$