What are the solutions to the equation 7x+(3x^2-4)=(5x-1)-(x-1) ?

The solutions to the equation 7x+(3x^2-4)=(5x-1)-(x-1) are x=(-3+sqrt(57))/6 ,x=(-3-sqrt(57))/6

Find the zeros of 7x+(3x^2-4)=(5x-1)-(x-1)

The zeros of 7x+(3x^2-4)=(5x-1)-(x-1) are x=(-3+sqrt(57))/6 ,x=(-3-sqrt(57))/6

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

x = \frac{- 3 + 57}{6}, x = \frac{- 3 - 57}{6}

Decimal Notation

x = 0.75830 \dots, x = - 1.75830 \dots

Solution steps

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

Expand

(5 x - 1) - (x - 1) : 4 x

7 x + 3 x^{2} - 4 = 4 x

Move

4 x

to the left side

3 x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

3^{2} - 4 \cdot 3 (- 4) = 57

x_{1, 2} = \frac{- 3 \pm 57}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 3 + 57}{2 \cdot 3}, x_{2} = \frac{- 3 - 57}{2 \cdot 3}

x = \frac{- 3 + 57}{2 \cdot 3} : \frac{- 3 + 57}{6}

x = \frac{- 3 - 57}{2 \cdot 3} : \frac{- 3 - 57}{6}

The solutions to the quadratic equation are:

x = \frac{- 3 + 57}{6}, x = \frac{- 3 - 57}{6}

0 = x^{2} - 18 x + 32

- 0.5 x^{2} + 1.05 x + 0.4 = - 8.5

6 = x^{2} + 5 x

27 t^{2} - 36 t + 12 = 0

\frac{x + 7}{6} = \frac{x ^{2} - 1}{3}

What are the solutions to the equation 7x+(3x^2-4)=(5x-1)-(x-1) ?
The solutions to the equation 7x+(3x^2-4)=(5x-1)-(x-1) are x=(-3+sqrt(57))/6 ,x=(-3-sqrt(57))/6
Find the zeros of 7x+(3x^2-4)=(5x-1)-(x-1)
The zeros of 7x+(3x^2-4)=(5x-1)-(x-1) are x=(-3+sqrt(57))/6 ,x=(-3-sqrt(57))/6