Solve 3x^2+17x-6=0 3x^2+18x-x-6=0 3x(x+6)-1(x+6)=0 (3x-1)(x+6)=0 3x-1=0orx+6=0 Based on the work shown to the left,which of these values are possible solutions of the equation? Check all of the boxes that apply. x=-6 x=6 x=-(1)/(3) Q x=(1)/(3) x=0

To solve the equation $3x^{2}+17x-6=0$, we can use factoring.<br /><br />Step 1: Factor the quadratic equation.<br />$3x^{2}+17x-6=0$<br />$3x^{2}+18x-x-6=0$<br />$3x(x+6)-1(x+6)=0$<br />$(3x-1)(x+6)=0$<br /><br />Step 2: Set each factor equal to zero and solve for x.<br />$3x-1=0$ or $x+6=0$<br />$x=\frac{1}{3}$ or $x=-6$<br /><br />Therefore, the possible solutions of the equation are:<br />- $x=-6$<br />- $x=\frac{1}{3}$<br /><br />So, the correct answers are:<br />- $x=-6$<br />- $x=\frac{1}{3}$