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

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

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

We move all terms to the left:

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


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

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

We add all the numbers together, and all the variables

F^2-3F-8

Back to the equation:

-(F^2-3F-8)


We get rid of parentheses

-F^2+F+3F+8=0

We add all the numbers together, and all the variables

-1F^2+4F+8=0

a = -1; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-1)·8
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*-1}=\frac{-4-4\sqrt{3}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*-1}=\frac{-4+4\sqrt{3}}{-2}
$