Solve 6x^2-13x=5

To be able to solve a quadratic, you must make it equal to zero. So the first thing you have to do is take away 5 from both sides. You could take away 6x^2-13x from each side, but it is much easier to keep your x^2 value positive. 6x^2-13x-5=0

Now look at the coefficients of each term (the number before the x), because the coefficient of x^2 is more than 1, we have to use the ac method to factorise this quadratic. Label each coefficient abc; a=6 b=-13 c=-5, it's very important to remember to keep the signs "stuck" to each coefficient.

Do ac=-30, you're now looking for two numbers that will multiply to make -30 (ac) and add together to make -13 (b), it can help to write out all the factors of -30. Factors of -30: 1,30 2,15 3,10 5,6 (one negative and one positive for each) -15x2=-30=ac -15+2=-13=b -15 and 2 are the two numbers you are looking for

Separate out the b term into the two different numbers that you have just found, now split your equation in half and factorise each side. You should get two numbers outside the brackets, and two of the same brackets. Put the two numbers outside into another bracket, and you have factorised the quadratic. 6x^2+2x-15x-5=0 2x(3x+1)-5(3x+1)=0 (2x-5)(3x+1)=0

To solve, make each bracket =0, and then solve for x, you should get two different values of x. Remember that you can check your work by expanding out the brackets, or by substituting in the values of x that you have found! 2x-5=0 3x+1=0 x=5/2 x=-1/3