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

The solutions to the equation 4x^2-3x+7=2x^2+x+7 are x=2,x=0

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

The zeros of 4x^2-3x+7=2x^2+x+7 are x=2,x=0

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

x = 2, x = 0

Solution steps

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

Move

7

to the left side

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

Move

x

to the left side

4 x^{2} - 4 x = 2 x^{2}

Move

2 x^{2}

to the left side

2 x^{2} - 4 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 \cdot 0 = 4

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

Separate the solutions

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

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

x = \frac{- ( - 4 ) - 4}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

- 3 (6 y - 4) \geq 2 (2 y - 5)

- 5 - r = 1 + 2 r

\frac{3}{2} \cdot 5

- 19 ∣ x ∣ + (4 x^{2} - 5) < 0

2 x^{2} + 8 x - 7 = - 2

What are the solutions to the equation 4x^2-3x+7=2x^2+x+7 ?
The solutions to the equation 4x^2-3x+7=2x^2+x+7 are x=2,x=0
Find the zeros of 4x^2-3x+7=2x^2+x+7
The zeros of 4x^2-3x+7=2x^2+x+7 are x=2,x=0