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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -((F^2+3F-4)), so:

(F^2+3F-4)

We get rid of parentheses

F^2+3F-4

Back to the equation:

-(F^2+3F-4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-3F+6=0

a = -1; b = -3; c = +6;
Δ = b2-4ac
Δ = -32-4·(-1)·6
Δ = 33
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{-(-3)-\sqrt{33}}{2*-1}=\frac{3-\sqrt{33}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*-1}=\frac{3+\sqrt{33}}{-2}
$