What are the solutions to the equation (x+5)(2x-3)=2(x+1) ?

The solutions to the equation (x+5)(2x-3)=2(x+1) are x=(-5+sqrt(161))/4 ,x=(-5-sqrt(161))/4

Find the zeros of (x+5)(2x-3)=2(x+1)

The zeros of (x+5)(2x-3)=2(x+1) are x=(-5+sqrt(161))/4 ,x=(-5-sqrt(161))/4

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(x + 5) (2 x - 3) = 2 (x + 1)

x = \frac{- 5 + 161}{4}, x = \frac{- 5 - 161}{4}

Decimal Notation

x = 1.92214 \dots, x = - 4.42214 \dots

Solution steps

(x + 5) (2 x - 3) = 2 (x + 1)

Expand

(x + 5) (2 x - 3) : 2 x^{2} + 7 x - 15

Expand

2 (x + 1) : 2 x + 2

2 x^{2} + 7 x - 15 = 2 x + 2

Move

2

to the left side

2 x^{2} + 7 x - 17 = 2 x

Move

2 x

to the left side

2 x^{2} + 5 x - 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 17 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 17) = 161

x_{1, 2} = \frac{- 5 \pm 161}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 161}{2 \cdot 2}, x_{2} = \frac{- 5 - 161}{2 \cdot 2}

x = \frac{- 5 + 161}{2 \cdot 2} : \frac{- 5 + 161}{4}

x = \frac{- 5 - 161}{2 \cdot 2} : \frac{- 5 - 161}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 161}{4}, x = \frac{- 5 - 161}{4}

(3 x - 2) (x + 1) = 8 x - 3

3 x^{2} + 5 x - 28 = 0

(1 + 5 x)^{2} = (4 x - 1) 2^{2} + 9 x (x - 2)

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

x, y = 3 x^{2} - 30 x + 72

What are the solutions to the equation (x+5)(2x-3)=2(x+1) ?
The solutions to the equation (x+5)(2x-3)=2(x+1) are x=(-5+sqrt(161))/4 ,x=(-5-sqrt(161))/4
Find the zeros of (x+5)(2x-3)=2(x+1)
The zeros of (x+5)(2x-3)=2(x+1) are x=(-5+sqrt(161))/4 ,x=(-5-sqrt(161))/4