What are the solutions to the equation (3x^2-13x+14)(x^2-4x+1)=0 ?

The solutions to the equation (3x^2-13x+14)(x^2-4x+1)=0 are x=2,x= 7/3 ,x=2+sqrt(3),x=2-sqrt(3)

Find the zeros of (3x^2-13x+14)(x^2-4x+1)=0

The zeros of (3x^2-13x+14)(x^2-4x+1)=0 are x=2,x= 7/3 ,x=2+sqrt(3),x=2-sqrt(3)

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(3 x^{2} - 13 x + 14) (x^{2} - 4 x + 1) = 0

x = 2, x = \frac{7}{3}, x = 2 + 3, x = 2 - 3

Decimal Notation

x = 2, x = 2.33333 \dots, x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

(3 x^{2} - 13 x + 14) (x^{2} - 4 x + 1) = 0

Factor

(3 x^{2} - 13 x + 14) (x^{2} - 4 x + 1) : (x - 2) (3 x - 7) (x^{2} - 4 x + 1)

(x - 2) (3 x - 7) (x^{2} - 4 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or 3 x - 7 = 0 or x^{2} - 4 x + 1 = 0

Solve

x - 2 = 0 : x = 2

Solve

3 x - 7 = 0 : x = \frac{7}{3}

Solve

x^{2} - 4 x + 1 = 0 : x = 2 + 3, x = 2 - 3

The solutions are

x = 2, x = \frac{7}{3}, x = 2 + 3, x = 2 - 3

y, x = (y + 7)^{3}

r^{3} + 3 r^{2} - 9 r - 27 = 0

v^{3} = 343

x, 6 y - 12 x^{3} = 0

x^{3} + x^{2} + x + 1 = 4

What are the solutions to the equation (3x^2-13x+14)(x^2-4x+1)=0 ?
The solutions to the equation (3x^2-13x+14)(x^2-4x+1)=0 are x=2,x= 7/3 ,x=2+sqrt(3),x=2-sqrt(3)
Find the zeros of (3x^2-13x+14)(x^2-4x+1)=0
The zeros of (3x^2-13x+14)(x^2-4x+1)=0 are x=2,x= 7/3 ,x=2+sqrt(3),x=2-sqrt(3)