What are the solutions to the equation 0= 1/116 (3x^2-18x+24) ?

The solutions to the equation 0= 1/116 (3x^2-18x+24) are x=2,x=4

Find the zeros of 0= 1/116 (3x^2-18x+24)

The zeros of 0= 1/116 (3x^2-18x+24) are x=2,x=4

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0 = \frac{1}{116} (3 x^{2} - 18 x + 24)

x = 2, x = 4

Solution steps

0 = \frac{1}{116} (3 x^{2} - 18 x + 24)

Switch sides

\frac{1}{116} (3 x^{2} - 18 x + 24) = 0

Factor

\frac{1}{116} (3 x^{2} - 18 x + 24) : \frac{3}{116} (x - 2) (x - 4)

\frac{3}{116} (x - 2) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x - 4 = 0

Solve

x - 2 = 0 : x = 2

Solve

x - 4 = 0 : x = 4

The solutions to the quadratic equation are:

x = 2, x = 4

36 = x^{2} + 5 x

0 = 6 x - x^{2}

6^{2} - 4^{2} = x^{2}

5 x^{2} + 8 x + 4 = 4 x^{2} + 6 x + 3

X^{2} - X - 20 = 0

What are the solutions to the equation 0= 1/116 (3x^2-18x+24) ?
The solutions to the equation 0= 1/116 (3x^2-18x+24) are x=2,x=4
Find the zeros of 0= 1/116 (3x^2-18x+24)
The zeros of 0= 1/116 (3x^2-18x+24) are x=2,x=4