What are the solutions to the equation x(2x-4)+6=3x^2-7x ?

The solutions to the equation x(2x-4)+6=3x^2-7x are x=-(-3+sqrt(33))/2 ,x=(3+sqrt(33))/2

Find the zeros of x(2x-4)+6=3x^2-7x

The zeros of x(2x-4)+6=3x^2-7x are x=-(-3+sqrt(33))/2 ,x=(3+sqrt(33))/2

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x (2 x - 4) + 6 = 3 x^{2} - 7 x

x = - \frac{- 3 + 33}{2}, x = \frac{3 + 33}{2}

Decimal Notation

x = - 1.37228 \dots, x = 4.37228 \dots

Solution steps

x (2 x - 4) + 6 = 3 x^{2} - 7 x

Expand

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

2 x^{2} - 4 x + 6 = 3 x^{2} - 7 x

Move

7 x

to the left side

2 x^{2} + 3 x + 6 = 3 x^{2}

Move

3 x^{2}

to the left side

- x^{2} + 3 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) \cdot 6}{2 ( - 1 )}

3^{2} - 4 (- 1) \cdot 6 = 33

x_{1, 2} = \frac{- 3 \pm 33}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 3 + 33}{2 ( - 1 )}, x_{2} = \frac{- 3 - 33}{2 ( - 1 )}

x = \frac{- 3 + 33}{2 ( - 1 )} : - \frac{- 3 + 33}{2}

x = \frac{- 3 - 33}{2 ( - 1 )} : \frac{3 + 33}{2}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 33}{2}, x = \frac{3 + 33}{2}

18 x^{2} = - 21 x

7.3 x + 1.4 + (- 5.9) x^{2} = 0

w^{2} = \frac{36}{100}

(y - 3)^{2} - 72 = 0

2 x^{2} + 11 x + 6 = 0

What are the solutions to the equation x(2x-4)+6=3x^2-7x ?
The solutions to the equation x(2x-4)+6=3x^2-7x are x=-(-3+sqrt(33))/2 ,x=(3+sqrt(33))/2
Find the zeros of x(2x-4)+6=3x^2-7x
The zeros of x(2x-4)+6=3x^2-7x are x=-(-3+sqrt(33))/2 ,x=(3+sqrt(33))/2