Question

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

Answer 1

A = -10 b = 12 c = -9

x = [-12 +- sq root (144 - 4*-10*-9)] / -20

x = [-12 +- sq root (144 -360)] / -20

x1 = 3/5 + (14.7 i / -20)

x2 = 3/5 - (14.7 i / -20)

Answer 2

Answer:

$x =\frac{-12+i\sqrt{216}}{-20},\frac{12+i\sqrt{216}}{20}$

Step-by-step explanation:

Given : $-10x^2 + 12x -9 = 0$

To Find: What are the roots of x?

Solution:

$-10x^2 + 12x -9 = 0$

We will solve this by quadratic formula :

Formula : $x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

General form of quadratic equation: $ax^2+bx+c=0$

On Comparing the given equation with general form.

a = -10

b= 12

c = -9

Substitute the values in the formula :

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

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

$x =\frac{-12\pm\sqrt{-216}}{-20}$

$x =\frac{-12+\sqrt{-216}}{-20},\frac{-12-\sqrt{-216}}{-20}$

$x =\frac{-12+i\sqrt{216}}{-20},\frac{-12-i\sqrt{216}}{-20}$

$x =\frac{-12+i\sqrt{216}}{-20},\frac{12+i\sqrt{216}}{20}$

Hence the roots of x are  $x =\frac{-12+i\sqrt{216}}{-20},\frac{12+i\sqrt{216}}{20}$