# how do i factor this using the quadratic equation 21x^2+22x-8=0

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We can also use the quadratic formula to factorize the equation:

21x^2+22x-8=0

x1 = [-b + sqrt(b^2 - 4ac)]/2a

We'll identify the coefficients a,b,c:

a = 21

b = 22

c = -8

b^2 - 4ac = 484 + 672

sqrt (b^2 - 4ac) = sqrt 1156

sqrt (b^2 - 4ac) = 34

x1 = (-22+34)/2*21

x1 = 12/42

x1 = 6/21

x2 = (-22-34)/2*21

x2 = -56/42

x2 = -28/21

We can now factorize the quadratic:

21x^2+22x-8 = 21(x - x1)(x - x2)

We'll substitute x1 and x2

**21x^2+22x-8 = 21(x - 6/21)(x + 28/21)**

**or**

**21x^2+22x-8 = (21x - 6)(21x + 28)**

The left hand side of the given equation is a quadratic expression. That is it involves use of powers of the variable x which is equal to 2, but not greater than 2.

This expression can be factored as follows.

21x^2 + 22x - 8

= 21x^2 + 28x - 6x - 8

= 7x(3x + 4) - 2(3x + 4)

= (3x + 4))(7x - 2)

Answer:

Factors of the quadratic equations are:

(3x + 4)) and (7x - 2)