# How do you solve the quadratic using the quadratic formula given 10x^2-11x+9=14x-6x^2 over the set of complex numbers?

Jun 26, 2017

$x = \frac{15}{32} - i \frac{\sqrt{351}}{32}$ or $\frac{15}{32} + i \frac{\sqrt{351}}{32}$

#### Explanation:

Quadratic formula gives the solution of quadratic equation $a {x}^{2} + b x + c = 0$ as

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

We have $10 {x}^{2} - 11 x + 9 = 14 x - 6 {x}^{2}$

or $10 {x}^{2} + 6 {x}^{2} - 11 x - 14 x + 9 = 0$

or $16 {x}^{2} - 15 x + 9 = 0$

and using quadratic formula

$x = \frac{- \left(- 15\right) \pm \sqrt{{\left(- 15\right)}^{2} - 4 \times 16 \times 9}}{2 \times 16}$

= $\frac{15 \pm \sqrt{225 - 576}}{32}$

= $\frac{15 \pm \sqrt{- 351}}{32}$

= $\frac{15}{32} \pm i \frac{\sqrt{351}}{32}$

i.e. $x = \frac{15}{32} - i \frac{\sqrt{351}}{32}$ or $\frac{15}{32} + i \frac{\sqrt{351}}{32}$