To find the solutions, set the equation equal to 0. You get [tex]4 x^{2} +34x = 0[/tex]. We are going to factor our things that both terms have in common, a 2, and an x. This gives you [tex]2x(2x+17)[/tex]. 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
