0=-5x^2+40x+1.2

Solution for 0=-5x^2+40x+1.2 equation:

0=-5x^2+40x+1.2

We move all terms to the left:

0-(-5x^2+40x+1.2)=0

We add all the numbers together, and all the variables

-(-5x^2+40x+1.2)=0

We get rid of parentheses

5x^2-40x-1.2=0

a = 5; b = -40; c = -1.2;
Δ = b2-4ac
Δ = -402-4·5·(-1.2)
Δ = 1624
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:

$\sqrt{\Delta}=\sqrt{1624}=\sqrt{4*406}=\sqrt{4}*\sqrt{406}=2\sqrt{406}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{406}}{2*5}=\frac{40-2\sqrt{406}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{406}}{2*5}=\frac{40+2\sqrt{406}}{10}
$