16-2f=3/2f+9=

Solution for 16-2f=3/2f+9= equation:

16-2f=3/2f+9=

We move all terms to the left:

16-2f-(3/2f+9)=0


Domain of the equation: 2f+9)!=0

f∈R


We get rid of parentheses

-2f-3/2f-9+16=0

We multiply all the terms by the denominator


-2f*2f-9*2f+16*2f-3=0

Wy multiply elements

-4f^2-18f+32f-3=0

We add all the numbers together, and all the variables

-4f^2+14f-3=0

a = -4; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·(-4)·(-3)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{37}}{2*-4}=\frac{-14-2\sqrt{37}}{-8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{37}}{2*-4}=\frac{-14+2\sqrt{37}}{-8}
$