2.5x^2-10x+2=0

Solution for 2.5x^2-10x+2=0 equation:

2.5x^2-10x+2=0

a = 2.5; b = -10; c = +2;
Δ = b2-4ac
Δ = -102-4·2.5·2
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{5}}{2*2.5}=\frac{10-4\sqrt{5}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{5}}{2*2.5}=\frac{10+4\sqrt{5}}{5}
$