4.905x^2-10x+1=0

Solution for 4.905x^2-10x+1=0 equation:

4.905x^2-10x+1=0

a = 4.905; b = -10; c = +1;
Δ = b2-4ac
Δ = -102-4·4.905·1
Δ = 80.38
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-\sqrt{80.38}}{2*4.905}=\frac{10-\sqrt{80.38}}{9.81}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{80.38}}{2*4.905}=\frac{10+\sqrt{80.38}}{9.81}
$