What are the solutions to the equation 2x^2=(4x-6x^2)+9-(6x+3) ?

The solutions to the equation 2x^2=(4x-6x^2)+9-(6x+3) are x=-1,x= 3/4

Find the zeros of 2x^2=(4x-6x^2)+9-(6x+3)

The zeros of 2x^2=(4x-6x^2)+9-(6x+3) are x=-1,x= 3/4

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2 x^{2} = (4 x - 6 x^{2}) + 9 - (6 x + 3)

x = - 1, x = \frac{3}{4}

Decimal Notation

x = - 1, x = 0.75

Solution steps

2 x^{2} = (4 x - 6 x^{2}) + 9 - (6 x + 3)

Expand

(4 x - 6 x^{2}) + 9 - (6 x + 3) : - 6 x^{2} - 2 x + 6

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

Switch sides

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

Move

2 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 8 ) \cdot 6}{2 ( - 8 )}

(- 2)^{2} - 4 (- 8) \cdot 6 = 14

x_{1, 2} = \frac{- ( - 2 ) \pm 14}{2 ( - 8 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 14}{2 ( - 8 )}, x_{2} = \frac{- ( - 2 ) - 14}{2 ( - 8 )}

x = \frac{- ( - 2 ) + 14}{2 ( - 8 )} : - 1

x = \frac{- ( - 2 ) - 14}{2 ( - 8 )} : \frac{3}{4}

The solutions to the quadratic equation are:

x = - 1, x = \frac{3}{4}

16 a^{2} - 8 a + 1 = 0

5 + 6 x = 2 x - 7

∣ 2 x + 4 ∣ = 14

x = - 2

∣ 2 x - 7 ∣ = 8

What are the solutions to the equation 2x^2=(4x-6x^2)+9-(6x+3) ?
The solutions to the equation 2x^2=(4x-6x^2)+9-(6x+3) are x=-1,x= 3/4
Find the zeros of 2x^2=(4x-6x^2)+9-(6x+3)
The zeros of 2x^2=(4x-6x^2)+9-(6x+3) are x=-1,x= 3/4