9/5f+32=68f

Solution for 9/5f+32=68f equation:

9/5f+32=68f

We move all terms to the left:

9/5f+32-(68f)=0


Domain of the equation: 5f!=0

f!=0/5

f!=0

f∈R


We add all the numbers together, and all the variables

-68f+9/5f+32=0

We multiply all the terms by the denominator


-68f*5f+32*5f+9=0

Wy multiply elements

-340f^2+160f+9=0

a = -340; b = 160; c = +9;
Δ = b2-4ac
Δ = 1602-4·(-340)·9
Δ = 37840
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:

$\sqrt{\Delta}=\sqrt{37840}=\sqrt{16*2365}=\sqrt{16}*\sqrt{2365}=4\sqrt{2365}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-4\sqrt{2365}}{2*-340}=\frac{-160-4\sqrt{2365}}{-680}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+4\sqrt{2365}}{2*-340}=\frac{-160+4\sqrt{2365}}{-680}
$