0.5x^2-14x+28=0

Solution for 0.5x^2-14x+28=0 equation:

0.5x^2-14x+28=0

a = 0.5; b = -14; c = +28;
Δ = b2-4ac
Δ = -142-4·0.5·28
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{35}}{2*0.5}=\frac{14-2\sqrt{35}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{35}}{2*0.5}=\frac{14+2\sqrt{35}}{1}
$