What are the solutions to the equation 3x^2-x+1=2x^2+5x-4 ?

The solutions to the equation 3x^2-x+1=2x^2+5x-4 are x=5,x=1

Find the zeros of 3x^2-x+1=2x^2+5x-4

The zeros of 3x^2-x+1=2x^2+5x-4 are x=5,x=1

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3 x^{2} - x + 1 = 2 x^{2} + 5 x - 4

x = 5, x = 1

Solution steps

3 x^{2} - x + 1 = 2 x^{2} + 5 x - 4

Move

4

to the left side

3 x^{2} - x + 5 = 2 x^{2} + 5 x

Move

5 x

to the left side

3 x^{2} - 6 x + 5 = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} - 6 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

x_{1, 2} = \frac{- ( - 6 ) \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 4}{2 \cdot 1}

x = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

x = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 5, x = 1

e^{- \frac{iπ}{3}}

k^{4} + 10 k^{3} + 35 k^{2} + 50 k + 24

x + 2 = 2 x - 1

9 (b + 7) = 72

(\frac{8}{9})^{3}

What are the solutions to the equation 3x^2-x+1=2x^2+5x-4 ?
The solutions to the equation 3x^2-x+1=2x^2+5x-4 are x=5,x=1
Find the zeros of 3x^2-x+1=2x^2+5x-4
The zeros of 3x^2-x+1=2x^2+5x-4 are x=5,x=1