What are the solutions to the equation 6x^2-4x+1=-9x+5 ?

The solutions to the equation 6x^2-4x+1=-9x+5 are x= 1/2 ,x=-4/3

Find the zeros of 6x^2-4x+1=-9x+5

The zeros of 6x^2-4x+1=-9x+5 are x= 1/2 ,x=-4/3

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6 x^{2} - 4 x + 1 = - 9 x + 5

x = \frac{1}{2}, x = - \frac{4}{3}

Decimal Notation

x = 0.5, x = - 1.33333 \dots

Solution steps

6 x^{2} - 4 x + 1 = - 9 x + 5

Move

5

to the left side

6 x^{2} - 4 x - 4 = - 9 x

Move

9 x

to the left side

6 x^{2} + 5 x - 4 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 6 (- 4) = 11

x_{1, 2} = \frac{- 5 \pm 11}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 5 + 11}{2 \cdot 6}, x_{2} = \frac{- 5 - 11}{2 \cdot 6}

x = \frac{- 5 + 11}{2 \cdot 6} : \frac{1}{2}

x = \frac{- 5 - 11}{2 \cdot 6} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{4}{3}

7 > - 2 x + 6

x^{2} + 4 x = 0

T + (\frac{3 T}{2}) + (\frac{( T + 2 )}{2 T}) = 4 T^{2}

(\frac{x ^{2}}{5})^{3}

x^{2} - 3 x y - 18 y^{2}

What are the solutions to the equation 6x^2-4x+1=-9x+5 ?
The solutions to the equation 6x^2-4x+1=-9x+5 are x= 1/2 ,x=-4/3
Find the zeros of 6x^2-4x+1=-9x+5
The zeros of 6x^2-4x+1=-9x+5 are x= 1/2 ,x=-4/3