250x^2-245x+49=0

Solution for 250x^2-245x+49=0 equation:

250x^2-245x+49=0

a = 250; b = -245; c = +49;
Δ = b2-4ac
Δ = -2452-4·250·49
Δ = 11025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11025}=105$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-245)-105}{2*250}=\frac{140}{500}
=7/25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-245)+105}{2*250}=\frac{350}{500}
=7/10
$