0.25x^2-8x+12=0

Solution for 0.25x^2-8x+12=0 equation:

0.25x^2-8x+12=0

a = 0.25; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·0.25·12
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{13}}{2*0.25}=\frac{8-2\sqrt{13}}{0.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{13}}{2*0.25}=\frac{8+2\sqrt{13}}{0.5}
$