3c+2(2c)+4(1/4-100)+1(1/4c)=7850

Solution for 3c+2(2c)+4(1/4-100)+1(1/4c)=7850 equation:

3c+2(2c)+4(1/4-100)+1(1/4c)=7850

We move all terms to the left:

3c+2(2c)+4(1/4-100)+1(1/4c)-(7850)=0


Domain of the equation: 4c)!=0

c!=0/1

c!=0

c∈R


determiningTheFunctionDomain
3c+22c+1(1/4c)-7850+4(1/4-100)=0

We add all the numbers together, and all the variables

3c+22c+1(+1/4c)-7850+4(1/4-100)=0

We add all the numbers together, and all the variables

25c+1(+1/4c)-7849=0

We multiply all the terms by the denominator


25c*4c)+1(-7849*4c)+1=0

Wy multiply elements

100c^2-31396c=0

a = 100; b = -31396; c = 0;
Δ = b2-4ac
Δ = -313962-4·100·0
Δ = 985708816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{985708816}=31396$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31396)-31396}{2*100}=\frac{0}{200}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31396)+31396}{2*100}=\frac{62792}{200}
=313+24/25
$