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

The solutions to the equation (x+2)(x-3)=2 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2

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

The zeros of (x+2)(x-3)=2 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2

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

x = \frac{1 + 33}{2}, x = \frac{1 - 33}{2}

Decimal Notation

x = 3.37228 \dots, x = - 2.37228 \dots

Solution steps

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

Expand

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

x^{2} - x - 6 = 2

Move

2

to the left side

x^{2} - x - 8 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 8) = 33

x_{1, 2} = \frac{- ( - 1 ) \pm 33}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 33}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 33}{2 \cdot 1}

x = \frac{- ( - 1 ) + 33}{2 \cdot 1} : \frac{1 + 33}{2}

x = \frac{- ( - 1 ) - 33}{2 \cdot 1} : \frac{1 - 33}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 33}{2}, x = \frac{1 - 33}{2}

15 = 2 x^{2} - 3 x + 1

6 x^{2} - 4 x - 13 = 0

x^{2} + (- 3) x = 0

3 x - 10 + x^{2} = 0

a^{2} = 8 a

What are the solutions to the equation (x+2)(x-3)=2 ?
The solutions to the equation (x+2)(x-3)=2 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2
Find the zeros of (x+2)(x-3)=2
The zeros of (x+2)(x-3)=2 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2