0=(0.2x^2)+1.2x-56

Solution for 0=(0.2x^2)+1.2x-56 equation:

0=(0.2x^2)+1.2x-56

We move all terms to the left:

0-((0.2x^2)+1.2x-56)=0

We add all the numbers together, and all the variables

-((0.2x^2)+1.2x-56)=0


We calculate terms in parentheses: -((0.2x^2)+1.2x-56), so:

(0.2x^2)+1.2x-56

We add all the numbers together, and all the variables

1.2x+(0.2x^2)-56

We get rid of parentheses

0.2x^2+1.2x-56

Back to the equation:

-(0.2x^2+1.2x-56)


We get rid of parentheses

-0.2x^2-1.2x+56=0

a = -0.2; b = -1.2; c = +56;
Δ = b2-4ac
Δ = -1.22-4·(-0.2)·56
Δ = 46.24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1.2)-\sqrt{46.24}}{2*-0.2}=\frac{1.2-\sqrt{46.24}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.2)+\sqrt{46.24}}{2*-0.2}=\frac{1.2+\sqrt{46.24}}{-0.4}
$