.2r2-7r+5=0

Solution for .2r2-7r+5=0 equation:

.2r^2-7r+5=0

a = .2; b = -7; c = +5;
Δ = b2-4ac
Δ = -72-4·.2·5
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3\sqrt{5}}{2*.2}=\frac{7-3\sqrt{5}}{0.4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3\sqrt{5}}{2*.2}=\frac{7+3\sqrt{5}}{0.4}
$