4.9x^2-20x+10=0

Solution for 4.9x^2-20x+10=0 equation:

4.9x^2-20x+10=0

a = 4.9; b = -20; c = +10;
Δ = b2-4ac
Δ = -202-4·4.9·10
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{51}}{2*4.9}=\frac{20-2\sqrt{51}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{51}}{2*4.9}=\frac{20+2\sqrt{51}}{9.8}
$