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

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

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

The zeros of x(2x+5)=4(x+7) are x= 7/2 ,x=-4

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

x = \frac{7}{2}, x = - 4

Solution steps

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

Expand

x (2 x + 5) : 2 x^{2} + 5 x

Expand

4 (x + 7) : 4 x + 28

2 x^{2} + 5 x = 4 x + 28

Move

28

to the left side

2 x^{2} + 5 x - 28 = 4 x

Move

4 x

to the left side

2 x^{2} + x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 28 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 28) = 15

x_{1, 2} = \frac{- 1 \pm 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 15}{2 \cdot 2}, x_{2} = \frac{- 1 - 15}{2 \cdot 2}

x = \frac{- 1 + 15}{2 \cdot 2} : \frac{7}{2}

x = \frac{- 1 - 15}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = - 4

10 x^{2} + 10 x^{2} + 1 = 0

7 n^{2} - 7 = 105

2 x^{2} - 4 x + 13

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

5 t^{2} + 3 t - 2 = 0

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