0.17x^2-20x+400=0

Solution for 0.17x^2-20x+400=0 equation:

0.17x^2-20x+400=0

a = 0.17; b = -20; c = +400;
Δ = b2-4ac
Δ = -202-4·0.17·400
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{2}}{2*0.17}=\frac{20-8\sqrt{2}}{0.34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{2}}{2*0.17}=\frac{20+8\sqrt{2}}{0.34}
$