select the approxiate value of x that is solutions for (x)=0 value f(x)=-3x^2+2x+8

Find the appropriate value(s) of x that is/are the solution(s) to f(x)=0 where f(x)=-3x^2+2x+8.

The values of x that make f(x) = 0 are called the Zeros of f(x). The Zeros are also the x-intercepts if they are real. If the Zeros are Imaginary then f(x) has no x-intercepts.

So we want to solve the Quadratic Equation 0=-3x^2+2x+8 for the x-values that make it true, its Zeros, a.k.a., Roots.

Let's use the Quadratic Formula to find the Zeros:

If 0 = ax^2 + bx + c, then

x = (- b ± sqrt(b^2-4ac))/(2a)

Identify parameters:

a = -3, the coefficient of x^2

b = 2, the coefficient of x

c = 8, the constant

Substitute:

x = (- 2 ± sqrt(2^2-4(-3)(8)))/(2(-3))

Simplify:

x = (- 2 ± sqrt(4+4(24)))/(-2(3))

The values of x that make f(x) = 0 are called the Zeros of f(x). The Zeros are also the x-intercepts if they are real. If the Zeros are Imaginary then f(x) has no x-intercepts.

So we want to solve the Quadratic Equation 0=-3x^2+2x+8 for the x-values that make it true, its Zeros, a.k.a., Roots.

Let's use the Quadratic Formula to find the Zeros:

If 0 = ax^2 + bx + c, then

x = (- b ± sqrt(b^2-4ac))/(2a)

Identify parameters:

a = -3, the coefficient of x^2

b = 2, the coefficient of x

c = 8, the constant

Substitute:

x = (- 2 ± sqrt(2^2-4(-3)(8)))/(2(-3))

Simplify:

x = (- 2 ± sqrt(4+4(24)))/(-2(3))

x = (- 2 ± sqrt(4(1+24)))/(-2(3))

x = (- 2 ± sqrt(4(25)))/(-2(3))

x = (- 2 ± 2*5)/(-2(3))

x = (1 ± 5)/3

x = -4/3 or 2