3/5c+9/10c=48

Solution for 3/5c+9/10c=48 equation:

3/5c+9/10c=48

We move all terms to the left:

3/5c+9/10c-(48)=0


Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R



Domain of the equation: 10c!=0

c!=0/10

c!=0

c∈R


We calculate fractions


30c/50c^2+45c/50c^2-48=0

We multiply all the terms by the denominator


30c+45c-48*50c^2=0

We add all the numbers together, and all the variables

75c-48*50c^2=0

Wy multiply elements

-2400c^2+75c=0

a = -2400; b = 75; c = 0;
Δ = b2-4ac
Δ = 752-4·(-2400)·0
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(75)-75}{2*-2400}=\frac{-150}{-4800}
=1/32
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(75)+75}{2*-2400}=\frac{0}{-4800}
=0
$