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