8+0.2x^2=28

Solution for 8+0.2x^2=28 equation:

8+0.2x^2=28

We move all terms to the left:

8+0.2x^2-(28)=0

We add all the numbers together, and all the variables

0.2x^2-20=0

a = 0.2; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·0.2·(-20)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*0.2}=\frac{-4}{0.4}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*0.2}=\frac{4}{0.4}
=10
$