do I do this correct and if not can you please show and tell me what I did wrong? thank you
6x^2-7x=3
We have to move the 3 to the left so it becomes
6x^2-7x-3=0
6x^2-7x-3=0 this makes this a quadratic equations 6*x^2-7*x-3=0 and has the answer of 1.5,-0 so
0=0
solution is x=1.5, x= -0.3333333333
I don't get the 0=0 step but your solution is correct for x = 1.5 and x = -0.3333.
so can you please show me what i am doing wrong with this?
I don't know that you did anything wrong. How did you solve the quadratic? Did you use the quadratic formula?
To solve the equation 6x^2 - 7x = 3, you have correctly moved the 3 to the left side to get 6x^2 - 7x - 3 = 0. However, the next step of finding the values of x is incorrect.
To solve the quadratic equation 6x^2 - 7x - 3 = 0, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a),
where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In this case, a = 6, b = -7, and c = -3. Plugging in these values into the quadratic formula, we have:
x = (-(-7) ± √((-7)^2 - 4 * 6 * (-3))) / (2 * 6)
= (7 ± √(49 + 72)) / 12
= (7 ± √121) / 12
= (7 ± 11) / 12.
So, the solutions to the quadratic equation are:
x = (7 + 11) / 12 = 1.5,
x = (7 - 11) / 12 = -0.33333... (or -1/3).
Therefore, the correct solution to the equation 6x^2 - 7x = 3 is x = 1.5 and x = -0.33333... (or x = -1/3).