Question

Select the two values of x that are roots of this equation. x2+3x-6+0

Answer 1

`\displaystyle x_1=(-3-√(31))/(2)` and
`\displaystyle x_2=(-3+√(31))/(2)`

Explanation:

I assume you mean
`x^2+3x-6=0` as your equation:

`\displaystyle x=(-b\pm√(b^2-4ac))/(2a)\\ \\x=(-3\pm√(3^2-4(1)(-6)))/(2(1))\\ \\x=(-3\pm√(9+24))/(2)\\ \\x=(-3\pm√(31))/(2)`

Thus,
`\displaystyle x_1=(-3-√(31))/(2)` and
`\displaystyle x_2=(-3+√(31))/(2)` are both roots to the equation.

Answer 2

x = ±sqrt(33)/2 - 3/2

Explanation:

Solve for x:

x^2 + 3 x - 6 = 0

Add 6 to both sides:

x^2 + 3 x = 6

Add 9/4 to both sides:

x^2 + 3 x + 9/4 = 33/4

Write the left hand side as a square:

(x + 3/2)^2 = 33/4

Take the square root of both sides:

x + 3/2 = sqrt(33)/2 or x + 3/2 = -sqrt(33)/2

Subtract 3/2 from both sides:

x = sqrt(33)/2 - 3/2 or x + 3/2 = -sqrt(33)/2

Subtract 3/2 from both sides:

Answer: x = ±sqrt(33)/2 - 3/2