Question

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

Answer 1

The two solutions are

`x_1 = 6/7 + i sqrt(34)/7`

`x_2 = 6/7 - i 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=7x^2-12x+10`

which of course is the same equation as

`7x^2-12x+10=0`

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

`x_{1,2} = \frac{-b\pm\sqrt{\Delta}}{2a}`

where `\Delta = b^2-4ac`

Let's compute `\Delta` serparately:

`b^2-4ac = (-12)^2-4*7*10 = 144-280 = -136`

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

`sqrt(-136) = sqrt(-1)*sqrt(136) = sqrt(-1)*sqrt(4*34) = sqrt(-1)*sqrt(4)*sqrt(34) = 2isqrt(34)`

So, the solving formula becomes

`x_{1,2} = \frac{12\pm2isqrt(34)}{14} = \frac{6\pmisqrt(34)}{7}`