Which are the solutions of the quadratic equation? x2 = 9x + 6 startfraction negative 9 minus startroot 105 endroot over 2 endfraction comma startfraction negative 9 + startroot 105 endroot over 2 endfraction startfraction negative 9 minus startroot 57 endroot over 2 endfraction comma startfraction negative 9 + startroot 57 endroot over 2 endfraction startfraction 9 minus startroot 105 endroot over 2 endfraction comma startfraction 9 + startroot 105 endroot over 2 endfraction startfraction 9 minus startroot 57 endroot over 2 endfraction comma startfraction 9 + startroot 57 endroot over 2 endfraction

The correct answer is C. 9 -√ 105/ 2,  9 + √ 105/ 2

Step-by-step explanation:

Let's find the solutions for the quadratic equation:

x² = 9x + 6

x² - 9x -6 = 0

Using the quadratic formula, we have:

x = [- (-9) +/- √ (-9)² - 4 * 1 * -6]/2 * 1

x = [ 9 +/- √ (81 + 24]/2

x = [ 9 +/- √ (105]/2

x₁  =  [ 9 + √ (105]/2

x₂ =  [ 9 - √ (105]/2

The correct answer is C. 9 -√ 105/ 2,  9 + √ 105/ 2

Note: Same answer than question 14133688, answered by me today.

x = 9+√105/2 or 9-√105/2

Step-by-step explanation:

Given the quadratic function x² = 9x+6

Re arranging we have;

x²-9x-6 = 0

Comparing this equation to general form ax²+bx+c = 0;

a = 1, b = -9, c = -6

Using the general formula to find x.

x = -b±√b²-4ac/2a

x = -(-9)±√81-4(1)(-6)/2

x = 9±√81+24/2

x = 9±√105/2

x = 9+√105/2 or 9-√105/2

mark other person brainliest

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.