ax^2+bx+c=0 ⇔ x=- b±sqrt(b^2-4ac)/2a
If we just want to know the number of solutions, and not the solutions themselves, we only need to work with the discriminant. Let's first write all terms on the left hand side of the equation.
6x^2+x-4=12 ⇔ 6x^2+x-16=0
Since the given equation is in standard form, we can identify the values of a, b, and c.
6x^2+ 1x+( - 16)=0
Finally, let's evaluate the discriminant.
Since the discriminant is 385, greater than zero, the quadratic equation has two real roots.
Extra
Further information
If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions.
