What are the solutions to the equation (3x-8)(x+2)=2x^2 ?

The solutions to the equation (3x-8)(x+2)=2x^2 are x=1+sqrt(17),x=1-sqrt(17)

Find the zeros of (3x-8)(x+2)=2x^2

The zeros of (3x-8)(x+2)=2x^2 are x=1+sqrt(17),x=1-sqrt(17)

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(3 x - 8) (x + 2) = 2 x^{2}

x = 1 + 17, x = 1 - 17

Decimal Notation

x = 5.12310 \dots, x = - 3.12310 \dots

Solution steps

(3 x - 8) (x + 2) = 2 x^{2}

Expand

(3 x - 8) (x + 2) : 3 x^{2} - 2 x - 16

3 x^{2} - 2 x - 16 = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} - 2 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 16 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 16) = 217

x_{1, 2} = \frac{- ( - 2 ) \pm 2 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 17}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 17}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 17}{2 \cdot 1} : 1 + 17

x = \frac{- ( - 2 ) - 2 17}{2 \cdot 1} : 1 - 17

The solutions to the quadratic equation are:

x = 1 + 17, x = 1 - 17

5 (x + 2)^{2} + 23 (x + 2) + 12 = 0

w^{2} = 6 w

7 a^{2} - 300 = 43

9 y^{2} - 12 y + 7 = 0

(5 x + 4) (x + 1) = 2

What are the solutions to the equation (3x-8)(x+2)=2x^2 ?
The solutions to the equation (3x-8)(x+2)=2x^2 are x=1+sqrt(17),x=1-sqrt(17)
Find the zeros of (3x-8)(x+2)=2x^2
The zeros of (3x-8)(x+2)=2x^2 are x=1+sqrt(17),x=1-sqrt(17)