What are the solutions to the equation 2x^2-3x-(6-4x)=0 ?

The solutions to the equation 2x^2-3x-(6-4x)=0 are x= 3/2 ,x=-2

Find the zeros of 2x^2-3x-(6-4x)=0

The zeros of 2x^2-3x-(6-4x)=0 are x= 3/2 ,x=-2

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2 x^{2} - 3 x - (6 - 4 x) = 0

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

Decimal Notation

x = 1.5, x = - 2

Solution steps

2 x^{2} - 3 x - (6 - 4 x) = 0

Expand

2 x^{2} - 3 x - (6 - 4 x) : 2 x^{2} + x - 6

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 6) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 2}, x_{2} = \frac{- 1 - 7}{2 \cdot 2}

x = \frac{- 1 + 7}{2 \cdot 2} : \frac{3}{2}

x = \frac{- 1 - 7}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

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

y (y - 1) = 56

x^{2} - 14 x + 130 = 0

2 x^{2} + 9 x = x - 14

(x^{2} + x + 10) = 3

x, z = x^{2} - x y + y^{2}

What are the solutions to the equation 2x^2-3x-(6-4x)=0 ?
The solutions to the equation 2x^2-3x-(6-4x)=0 are x= 3/2 ,x=-2
Find the zeros of 2x^2-3x-(6-4x)=0
The zeros of 2x^2-3x-(6-4x)=0 are x= 3/2 ,x=-2