What are the solutions to the equation (2x+5)(x-2)=34 ?

The solutions to the equation (2x+5)(x-2)=34 are x=(-1+sqrt(353))/4 ,x=(-1-sqrt(353))/4

Find the zeros of (2x+5)(x-2)=34

The zeros of (2x+5)(x-2)=34 are x=(-1+sqrt(353))/4 ,x=(-1-sqrt(353))/4

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(2 x + 5) (x - 2) = 34

x = \frac{- 1 + 353}{4}, x = \frac{- 1 - 353}{4}

Decimal Notation

x = 4.44707 \dots, x = - 4.94707 \dots

Solution steps

(2 x + 5) (x - 2) = 34

Expand

(2 x + 5) (x - 2) : 2 x^{2} + x - 10

2 x^{2} + x - 10 = 34

Move

34

to the left side

2 x^{2} + x - 44 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 44 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 44) = 353

x_{1, 2} = \frac{- 1 \pm 353}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 353}{2 \cdot 2}, x_{2} = \frac{- 1 - 353}{2 \cdot 2}

x = \frac{- 1 + 353}{2 \cdot 2} : \frac{- 1 + 353}{4}

x = \frac{- 1 - 353}{2 \cdot 2} : \frac{- 1 - 353}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 353}{4}, x = \frac{- 1 - 353}{4}

4 x^{2} + 24 = 9

45 n^{2} + 10 n = 0

4 x^{2} = 25

3 x^{2} = - 4 x + 15

x^{2} - 6 x - 8 = - 1

What are the solutions to the equation (2x+5)(x-2)=34 ?
The solutions to the equation (2x+5)(x-2)=34 are x=(-1+sqrt(353))/4 ,x=(-1-sqrt(353))/4
Find the zeros of (2x+5)(x-2)=34
The zeros of (2x+5)(x-2)=34 are x=(-1+sqrt(353))/4 ,x=(-1-sqrt(353))/4