How do you solve #2x^2=x+28 #?
Move all values and variables to the left and either factor, solve with the quadratic equation, or solve by completing the square; you could also leave the expression alone and solve by graphing.
Next, try to factor this expression.
The factors of -28 are:
Of these factors, only the last pair will allow you to make
Check your work by plugging these values back into the original expression.
You could solve this quadratic equation by completing the square.
To do that, start by getting your quadratic to the general form
#color(blue)(x^2 + b/ax = -c/a)#
Now, you need to add a term to both sides of the equation so that the left side of the equation can be written as the square of a binomial.
The coefficient of the
The left side of the equation can now be rewritten as
You now have
Take the square root of both sides of the equation
The two solutions to the equation will thus be
I use the new Transforming Method.
Factor pairs of (-56) --> (-4, 14)(-7, 8). This sum is 1 = -b. Two real roots of (2) are: -7 and 8.
The 2 real roots of equation (1) are: