25c=9/5c+32

Solution for 25c=9/5c+32 equation:

25c=9/5c+32

We move all terms to the left:

25c-(9/5c+32)=0


Domain of the equation: 5c+32)!=0

c∈R


We get rid of parentheses

25c-9/5c-32=0

We multiply all the terms by the denominator


25c*5c-32*5c-9=0

Wy multiply elements

125c^2-160c-9=0

a = 125; b = -160; c = -9;
Δ = b2-4ac
Δ = -1602-4·125·(-9)
Δ = 30100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{30100}=\sqrt{100*301}=\sqrt{100}*\sqrt{301}=10\sqrt{301}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-10\sqrt{301}}{2*125}=\frac{160-10\sqrt{301}}{250}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+10\sqrt{301}}{2*125}=\frac{160+10\sqrt{301}}{250}
$