What are the solutions to the equation 7x^2-6x-4=-x+4x^2 ?

The solutions to the equation 7x^2-6x-4=-x+4x^2 are x=(5+sqrt(73))/6 ,x=(5-sqrt(73))/6

Find the zeros of 7x^2-6x-4=-x+4x^2

The zeros of 7x^2-6x-4=-x+4x^2 are x=(5+sqrt(73))/6 ,x=(5-sqrt(73))/6

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

x = \frac{5 + 73}{6}, x = \frac{5 - 73}{6}

Decimal Notation

x = 2.25733 \dots, x = - 0.59066 \dots

Solution steps

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

Move

4 x^{2}

to the left side

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

Move

x

to the left side

3 x^{2} - 5 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 4) = 73

x_{1, 2} = \frac{- ( - 5 ) \pm 73}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 73}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 73}{2 \cdot 3}

x = \frac{- ( - 5 ) + 73}{2 \cdot 3} : \frac{5 + 73}{6}

x = \frac{- ( - 5 ) - 73}{2 \cdot 3} : \frac{5 - 73}{6}

The solutions to the quadratic equation are:

x = \frac{5 + 73}{6}, x = \frac{5 - 73}{6}

x^{2} - 6 x + 7 = 0

4 x^{2} + 4 x = 9

7 x^{2} - 38 x - 24 = 0

x, x^{2} + y^{2} + x y = 3

2 x^{2} + 7 x + 11 = 0

What are the solutions to the equation 7x^2-6x-4=-x+4x^2 ?
The solutions to the equation 7x^2-6x-4=-x+4x^2 are x=(5+sqrt(73))/6 ,x=(5-sqrt(73))/6
Find the zeros of 7x^2-6x-4=-x+4x^2
The zeros of 7x^2-6x-4=-x+4x^2 are x=(5+sqrt(73))/6 ,x=(5-sqrt(73))/6