F(x)=^4-3x^2+5x-3

Solution for F(x)=^4-3x^2+5x-3 equation:

(F)=^4-3F^2+5F-3

We move all terms to the left:

(F)-(^4-3F^2+5F-3)=0

We get rid of parentheses

3F^2-5F+F+3-^4=0

We add all the numbers together, and all the variables

3F^2-4F=0

a = 3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·3·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*3}=\frac{0}{6}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*3}=\frac{8}{6}
=1+1/3
$