how do i work this problem? Solve by factoring: 6x2 + x = 5 {1, 5} {-1, 5/6} {-1, 5} {1, 5/6}
`1.` Subtract 5 from both sides `2.` Find two numbers that multiply to get \(ac\), yet add to get \(b\) `3.` Split the middle term using those two numbers `4.` Factor by grouping `5.` Notice the common term that remains. Factor that out as well `6.` Use zero product property to find x `1.` \(6x^2 + x - 5 = 0\) `2.` \(6 \times -5 = -30\) \(6 - 5 = 1\) `3.` \(6x^2 + 6x - 5x - 5\) `4.` \(6x(x + 1) - 5(x + 1)\) `5.` x + 1 is still common to both terms so factoring it out results in: \((x+1)(6x - 5)\) `6.` Set the above equal to zero, and solve for x: \((x + 1)(6x - 5) = 0\) \(x + 1 = 0\) \(6x - 5 = 0\) I'll leave that part for you to finish solving.
Join our real-time social learning platform and learn together with your friends!