F(x)=-0,04x^2+0,6x+1,8

Solution for F(x)=-0,04x^2+0,6x+1,8 equation:

(F)=-0.04F^2+0.6F+1.8

We move all terms to the left:

(F)-(-0.04F^2+0.6F+1.8)=0

We get rid of parentheses

0.04F^2-0.6F+F-1.8=0

We add all the numbers together, and all the variables

0.04F^2+0.4F-1.8=0

a = 0.04; b = 0.4; c = -1.8;
Δ = b2-4ac
Δ = 0.42-4·0.04·(-1.8)
Δ = 0.448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.4)-\sqrt{0.448}}{2*0.04}=\frac{-0.4-\sqrt{0.448}}{0.08}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.448}}{2*0.04}=\frac{-0.4+\sqrt{0.448}}{0.08}
$