To find the solutions, set the equation equal to 0. You get
![4 x^{2} +34x = 0](https://tex.z-dn.net/?f=4%20x%5E%7B2%7D%20%2B34x%20%3D%200)
. We are going to factor our things that both terms have in common, a 2, and an x. This gives you
![2x(2x+17)](https://tex.z-dn.net/?f=2x%282x%2B17%29)
. How do you know this is true? If you distribute back out, you get the same as the original problem.
Now set these both equal to zero. 2x = 0, 2x + 17 = 0. Divide both sides by 2, you get one solultion of x = 0. For the other solution, subtract 17 from both sides; 2x = - 17. Divide by 2, x = -17/2
Your solutions are x = {- 17/2, 0}
To check, plug these back into the original equation to see if they equal zero. 4(0)^2 + 34(0) = 0 + 0 = 0
4(- 17/2)^2 + 34(- 17/2) = 289 + (- 289) = 289 - 289 = 0