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

The solutions to the equation (2x+3)(x-2)=2 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4

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

The zeros of (2x+3)(x-2)=2 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4

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

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

Decimal Notation

x = 2.26556 \dots, x = - 1.76556 \dots

Solution steps

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

Expand

(2 x + 3) (x - 2) : 2 x^{2} - x - 6

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

Move

2

to the left side

2 x^{2} - x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 8 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 8) = 65

x_{1, 2} = \frac{- ( - 1 ) \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 65}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 65}{2 \cdot 2}

x = \frac{- ( - 1 ) + 65}{2 \cdot 2} : \frac{1 + 65}{4}

x = \frac{- ( - 1 ) - 65}{2 \cdot 2} : \frac{1 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

(x - 32)^{2} = 56

x (3 x - 7) = 6

(2 x + 1)^{2} - x^{2} + 5 x - 10 = 0

64 a z^{2} - 81 a^{3} = 0

- 0.005 x^{2} + 0.101 x - 0.193 = 0

What are the solutions to the equation (2x+3)(x-2)=2 ?
The solutions to the equation (2x+3)(x-2)=2 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4
Find the zeros of (2x+3)(x-2)=2
The zeros of (2x+3)(x-2)=2 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4