Question

What are the roots of x in -10x2 + 12x − 9 = 0?

Answer 1

Answer:

$x_1=\frac{3(2+i\sqrt{6})}{10}$

$x_2=\frac{3(2-i\sqrt{6})}{10}$

Step-by-step explanation:

Use the Quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

You can identify that, in this case:

$a=-10\\b=12\\c=-9$

Now you need to substitute these values into the formula:

$x=\frac{-12\±\sqrt{12^2-4(-10)(-9)}}{2(-10)}$

$x=\frac{-12\±\sqrt{-216}}{-20}$

Remember that:

$i=\sqrt{-1}$

Therefore,rewriting and simplifying, you get:

$x=\frac{-12\±6i\sqrt{6}}{-20}$

$x=\frac{-6(2\±i\sqrt{6})}{-2(10)}$

$x=\frac{3(2\±i\sqrt{6})}{10}$

Then, you get the following roots:

$x_1=\frac{3(2+i\sqrt{6})}{10}$

$x_2=\frac{3(2-i\sqrt{6})}{10}$

Answer 2

Answer:

x = 3[2 + i√-6]/10 or x = 3[2 - i√-6]/10

Step-by-step explanation:

Formula

Solution of a quadratic equation ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/2a

To find the solution

It is given that,

-10x² + 12x - 9 = 0

here a = -10, b = 12 and c = -9

x = [-b ± √(b² - 4ac)]/2a

  = [-12 ± √(12² - 4* -10 * -9)]/2 * -10

 = [-12 ± √(144 - 360)]/-20

 =[12 ± √-216]/20

 = [12 ± i6√-6]/20

 =  [6 ± i3√-6]/10

 = 3[2 ± i√-6]/10

x = 3[2 + i√-6]/10 or x = 3[2 - i√-6]/10