Answer:
[tex]x =\frac{-12+i\sqrt{216}}{-20},\frac{12+i\sqrt{216}}{20}[/tex]
Step-by-step explanation:
Given : [tex]-10x^2 + 12x -9 = 0[/tex]
To Find: What are the roots of x?
Solution:
[tex]-10x^2 + 12x -9 = 0[/tex]
We will solve this by quadratic formula :
Formula : [tex]x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
General form of quadratic equation: [tex]ax^2+bx+c=0[/tex]
On Comparing the given equation with general form.
a = -10
b= 12
c = -9
Substitute the values in the formula :
[tex]x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x =\frac{-12\pm\sqrt{12^2-4(-10)(-9)}}{2(-10)}[/tex]
[tex]x =\frac{-12\pm\sqrt{-216}}{-20}[/tex]
[tex]x =\frac{-12+\sqrt{-216}}{-20},\frac{-12-\sqrt{-216}}{-20}[/tex]
[tex]x =\frac{-12+i\sqrt{216}}{-20},\frac{-12-i\sqrt{216}}{-20}[/tex]
[tex]x =\frac{-12+i\sqrt{216}}{-20},\frac{12+i\sqrt{216}}{20}[/tex]
Hence the roots of x are [tex]x =\frac{-12+i\sqrt{216}}{-20},\frac{12+i\sqrt{216}}{20}[/tex]
