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

The solutions to the equation (3x-1)(x+2)-x(x-4)=0 are x=(-9+sqrt(97))/4 ,x=(-9-sqrt(97))/4

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

The zeros of (3x-1)(x+2)-x(x-4)=0 are x=(-9+sqrt(97))/4 ,x=(-9-sqrt(97))/4

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(3 x - 1) (x + 2) - x (x - 4) = 0

x = \frac{- 9 + 97}{4}, x = \frac{- 9 - 97}{4}

Decimal Notation

x = 0.21221 \dots, x = - 4.71221 \dots

Solution steps

(3 x - 1) (x + 2) - x (x - 4) = 0

Expand

(3 x - 1) (x + 2) - x (x - 4) : 2 x^{2} + 9 x - 2

2 x^{2} + 9 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

9^{2} - 4 \cdot 2 (- 2) = 97

x_{1, 2} = \frac{- 9 \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 9 + 97}{2 \cdot 2}, x_{2} = \frac{- 9 - 97}{2 \cdot 2}

x = \frac{- 9 + 97}{2 \cdot 2} : \frac{- 9 + 97}{4}

x = \frac{- 9 - 97}{2 \cdot 2} : \frac{- 9 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{- 9 + 97}{4}, x = \frac{- 9 - 97}{4}

40 = y^{2}

n^{2} - 5 n - 6 = 0

1.25 x^{2} - 25 x + 5 = 0

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

y^{2} = - 5

What are the solutions to the equation (3x-1)(x+2)-x(x-4)=0 ?
The solutions to the equation (3x-1)(x+2)-x(x-4)=0 are x=(-9+sqrt(97))/4 ,x=(-9-sqrt(97))/4
Find the zeros of (3x-1)(x+2)-x(x-4)=0
The zeros of (3x-1)(x+2)-x(x-4)=0 are x=(-9+sqrt(97))/4 ,x=(-9-sqrt(97))/4