5=1+((2x^2)/4)

Solution for 5=1+((2x^2)/4) equation:

5=1+((2x^2)/4)

We move all terms to the left:

5-(1+((2x^2)/4))=0

We multiply all the terms by the denominator


-(1+(2x^2+5*4))=0


We calculate terms in parentheses: -(1+(2x^2+5*4)), so:

1+(2x^2+5*4)

determiningTheFunctionDomain
(2x^2+5*4)+1

We get rid of parentheses

2x^2+1+5*4

We add all the numbers together, and all the variables

2x^2+21

Back to the equation:

-(2x^2+21)


We get rid of parentheses

-2x^2-21=0

a = -2; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·(-2)·(-21)
Δ = -168
Delta is less than zero, so there is no solution for the equation