What are the solutions to the equation 2x^2-7x+9=(x-3)(x+1)+3x ?

The solutions to the equation 2x^2-7x+9=(x-3)(x+1)+3x are x=6,x=2

Find the zeros of 2x^2-7x+9=(x-3)(x+1)+3x

The zeros of 2x^2-7x+9=(x-3)(x+1)+3x are x=6,x=2

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2 x^{2} - 7 x + 9 = (x - 3) (x + 1) + 3 x

x = 6, x = 2

Solution steps

2 x^{2} - 7 x + 9 = (x - 3) (x + 1) + 3 x

Expand

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

2 x^{2} - 7 x + 9 = x^{2} + x - 3

Move

3

to the left side

2 x^{2} - 7 x + 12 = x^{2} + x

Move

x

to the left side

2 x^{2} - 8 x + 12 = x^{2}

Move

x^{2}

to the left side

x^{2} - 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 12 = 4

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

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 1}

x = \frac{- ( - 8 ) + 4}{2 \cdot 1} : 6

x = \frac{- ( - 8 ) - 4}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 6, x = 2

2 x (x + 3) + 1 = 0

(x + 2)^{3} = x (x^{2} - 10 x + 25)

5 x^{2} + 20 x + 19 = 0

9 x^{2} - 125 = 0

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

What are the solutions to the equation 2x^2-7x+9=(x-3)(x+1)+3x ?
The solutions to the equation 2x^2-7x+9=(x-3)(x+1)+3x are x=6,x=2
Find the zeros of 2x^2-7x+9=(x-3)(x+1)+3x
The zeros of 2x^2-7x+9=(x-3)(x+1)+3x are x=6,x=2