What is the answer to (2x^2+6x-8)/(x+4)=0 ?

The answer to (2x^2+6x-8)/(x+4)=0 is x=1

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(2x^2+6x-8)/(x+4)=0

\frac{2 x ^{2} + 6 x - 8}{x + 4} = 0

x = 1

Solution steps

\frac{2 x ^{2} + 6 x - 8}{x + 4} = 0

Simplify

\frac{2 x ^{2} + 6 x - 8}{x + 4} : 2 (x - 1)

2 (x - 1) = 0

Divide both sides by

2

x - 1 = 0

Move

1

to the right side

x = 1

Verify Solutions

Find undefined (singularity) points:

x = - 4

Combine undefined points with solutions:

x = 1

y, x = \frac{1}{y} - 2

\frac{2 x + 2}{x - 2} = 5

\frac{2 x - 6}{x} = \frac{x - 3}{3}

\frac{1}{x} + 5 = \frac{1}{2 x}

3 = \frac{1 - x}{1 + x}

What is the answer to (2x^2+6x-8)/(x+4)=0 ?
The answer to (2x^2+6x-8)/(x+4)=0 is x=1
Solve for x: (2x^2+6x-8)/(x+4)=0
The solution is x=1