What are the solutions to the equation (2x+4)(x-7)=2(5x-3) ?

The solutions to the equation (2x+4)(x-7)=2(5x-3) are x=11,x=-1

Find the zeros of (2x+4)(x-7)=2(5x-3)

The zeros of (2x+4)(x-7)=2(5x-3) are x=11,x=-1

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(2 x + 4) (x - 7) = 2 (5 x - 3)

x = 11, x = - 1

Solution steps

(2 x + 4) (x - 7) = 2 (5 x - 3)

Expand

(2 x + 4) (x - 7) : 2 x^{2} - 10 x - 28

Expand

2 (5 x - 3) : 10 x - 6

2 x^{2} - 10 x - 28 = 10 x - 6

Move

6

to the left side

2 x^{2} - 10 x - 22 = 10 x

Move

10 x

to the left side

2 x^{2} - 20 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 ( - 22 )}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 (- 22) = 24

x_{1, 2} = \frac{- ( - 20 ) \pm 24}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 24}{2 \cdot 2}, x_{2} = \frac{- ( - 20 ) - 24}{2 \cdot 2}

x = \frac{- ( - 20 ) + 24}{2 \cdot 2} : 11

x = \frac{- ( - 20 ) - 24}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = 11, x = - 1

x, 0 = 2 x^{2} + y^{2}

15 x^{2} - 9 x + 12 = 0

5 x^{2} + 11 x - 1 = 0

x^{2} + 4 x = 0

- x^{2} + 8 x - 5

What are the solutions to the equation (2x+4)(x-7)=2(5x-3) ?
The solutions to the equation (2x+4)(x-7)=2(5x-3) are x=11,x=-1
Find the zeros of (2x+4)(x-7)=2(5x-3)
The zeros of (2x+4)(x-7)=2(5x-3) are x=11,x=-1