.04x^2+0.5x-0.2=2.4

Solution for .04x^2+0.5x-0.2=2.4 equation:

.04x^2+0.5x-0.2=2.4

We move all terms to the left:

.04x^2+0.5x-0.2-(2.4)=0

We add all the numbers together, and all the variables

.04x^2+0.5x-2.6=0

a = .04; b = 0.5; c = -2.6;
Δ = b2-4ac
Δ = 0.52-4·.04·(-2.6)
Δ = 0.666
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{-(0.5)-\sqrt{0.666}}{2*.04}=\frac{-0.5-\sqrt{0.666}}{0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.666}}{2*.04}=\frac{-0.5+\sqrt{0.666}}{0.08}
$