What are the solutions to the equation 2x^2-3x-3=2 ?

The solutions to the equation 2x^2-3x-3=2 are x= 5/2 ,x=-1

Find the zeros of 2x^2-3x-3=2

The zeros of 2x^2-3x-3=2 are x= 5/2 ,x=-1

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

x = \frac{5}{2}, x = - 1

Solution steps

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

Move

2

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 5) = 7

x_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 7}{2 \cdot 2}

x = \frac{- ( - 3 ) + 7}{2 \cdot 2} : \frac{5}{2}

x = \frac{- ( - 3 ) - 7}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - 1

9 (2 x + 8)^{2} + 41 = 42

- 2 x^{2} - 4 x + 1 = 0

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

7 x^{2} - 105 = 0

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

What are the solutions to the equation 2x^2-3x-3=2 ?
The solutions to the equation 2x^2-3x-3=2 are x= 5/2 ,x=-1
Find the zeros of 2x^2-3x-3=2
The zeros of 2x^2-3x-3=2 are x= 5/2 ,x=-1