How do you solve the quadratic with complex numbers given -7x^2+12x=10?

Oct 1, 2016

The two solutions are

${x}_{1} = \frac{6}{7} + i \frac{\sqrt{34}}{7}$

${x}_{2} = \frac{6}{7} - i \frac{\sqrt{34}}{7}$

Explanation:

First of all, let's bring everything to one side of the equation. I usually like to have the ${x}^{2}$ coefficient positive, so I'll move everything to the right side, obtaining

$0 = 7 {x}^{2} - 12 x + 10$

which of course is the same equation as

$7 {x}^{2} - 12 x + 10 = 0$

Now, let's set the usual formula for solving a quadratic equation:

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{\setminus \Delta}}{2 a}$

where $\setminus \Delta = {b}^{2} - 4 a c$

Let's compute $\setminus \Delta$ serparately:

${b}^{2} - 4 a c = {\left(- 12\right)}^{2} - 4 \cdot 7 \cdot 10 = 144 - 280 = - 136$

To compute the square root of $- 136$, we need complex numbers. You can observe that

$\sqrt{- 136} = \sqrt{- 1} \cdot \sqrt{136} = \sqrt{- 1} \cdot \sqrt{4 \cdot 34} = \sqrt{- 1} \cdot \sqrt{4} \cdot \sqrt{34} = 2 i \sqrt{34}$

So, the solving formula becomes

${x}_{1 , 2} = \setminus \frac{12 \setminus \pm 2 i \sqrt{34}}{14} = \setminus \frac{6 \setminus \pm i \sqrt{34}}{7}$