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There are several different methods for solving quadratic equations. The method you use depends on the equation in question.
Usually, the easiest method is by factorising the quadratic, so you want to check if you can do this first. Write the quadratic as
Ax2 + Bx + C = 0
Then consider the expression
(Ax + )(x + ) = 0
Where spaces have been left for real numbers. This will clearly multiply to give you the term in x2 so now the challenge is to find a pair of numbers that multiply together to give C and will multiply with the term in x to give B.
It might help to factorise C - which numbers can multiply together to give you C? Then trialling these numbers and seeing what coefficient they give you for x will enable you to find the factorisation, if it exists.
2x2 + 7x + 6 = 0
Write as (2x + ) (x + )
Note that numbers that will multiply to give us 6 are (6, 1), and (2, 3). Trying 6 and 1 quickly shows these numbers will not work, and we can see that:
(2x + 3 )(x + 2) = 0
when multiplied out gives the appropriate coefficient of 7 for x. The solutions are then the values of x that can make the contents of the brackets equal zero:
x = - 1.5 and x = -2.
The quadratic formula works regardless of whether or not the quadratic can be factorised. However, it is algebraically more complicated, so factorising is preferable if you can do it quickly. The quadratic formula gives the solutions as:
x = (- B + (B2 - 4AC)0.5 )/ 2A
x = (- B - (B2 - 4AC)0.5 )/ 2A
where A, B and C are the coefficients of a quadratic Ax2 + Bx + C = 0.
Note that if B2 - 4AC is less than zero, this involves taking the square root of a negative number, and so the quadratic will have no real solutions. However, complex solutions - mixtures of real and imaginary numbers - will exist.