0.5x^2+6x+4=20

Solution for 0.5x^2+6x+4=20 equation:

0.5x^2+6x+4=20

We move all terms to the left:

0.5x^2+6x+4-(20)=0

We add all the numbers together, and all the variables

0.5x^2+6x-16=0

a = 0.5; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·0.5·(-16)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{17}}{2*0.5}=\frac{-6-2\sqrt{17}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{17}}{2*0.5}=\frac{-6+2\sqrt{17}}{1}
$