# How do you solve  21x^2 + 22x - 24 = 0?

Feb 26, 2017

$x = - \frac{12}{7}$ or $x = \frac{2}{3}$

#### Explanation:

If the discriminant of a quadratic equation $a {x}^{2} + b x + c = 0$, which is $\Delta = {b}^{2} - 4 a c$ is a complete square of a rational number, we can solve it by factorization.

Here in the given equation $21 {x}^{2} + 22 x - 24 = 0$, $\Delta = {22}^{2} - 4 \times 21 \times \left(- 24\right) = 484 + 2016 = 2500 = {50}^{2}$, hence we can solve it by factorization.

For factorization, split middle term $b = 22$ in two parts whose product is $a c = - 21 \times 24 = - 504$. In other words two numbers, whose difference is $22$ and product is $504$. These are are $- 14$ and $36$.

Therefore, $21 {x}^{2} + 22 x - 24 = 0$ can be written as

$21 {x}^{2} - 14 x + 36 x - 24 = 0$

or $7 x \left(3 x - 2\right) + 12 \left(3 x - 2\right) = 0$

or $\left(7 x + 12\right) \left(3 x - 2\right) = 0$

Hence either $7 x + 12 = 0$ i.e. $x = - \frac{12}{7}$

or $3 x - 2 = 0$ i.e. $x = \frac{2}{3}$