F(x)=3/2x-3-5

Solution for F(x)=3/2x-3-5 equation:

(F)=3/2F-3-5

We move all terms to the left:

(F)-(3/2F-3-5)=0


Domain of the equation: 2F-3-5)!=0

We move all terms containing F to the left, all other terms to the right

2F-5)!=3

F∈R


We add all the numbers together, and all the variables

F-(3/2F-8)=0

We get rid of parentheses

F-3/2F+8=0

We multiply all the terms by the denominator


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

Wy multiply elements

2F^2+16F-3=0

a = 2; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·2·(-3)
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{70}}{2*2}=\frac{-16-2\sqrt{70}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{70}}{2*2}=\frac{-16+2\sqrt{70}}{4}
$