What are the solutions to the equation -2x^2-3x+9=0 ?

The solutions to the equation -2x^2-3x+9=0 are x=-3,x= 3/2

Find the zeros of -2x^2-3x+9=0

The zeros of -2x^2-3x+9=0 are x=-3,x= 3/2

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- 2 x^{2} - 3 x + 9 = 0

x = - 3, x = \frac{3}{2}

Decimal Notation

x = - 3, x = 1.5

Solution steps

- 2 x^{2} - 3 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 9}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 9 = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 9}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 9}{2 ( - 2 )} : - 3

x = \frac{- ( - 3 ) - 9}{2 ( - 2 )} : \frac{3}{2}

The solutions to the quadratic equation are:

x = - 3, x = \frac{3}{2}

x^{2} + 7 = 2 x

5 b^{2} - 36 b = 32

(2 y - 1)^{2} - (2 y - 1) y + y^{2} - 7 = 0

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

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

What are the solutions to the equation -2x^2-3x+9=0 ?
The solutions to the equation -2x^2-3x+9=0 are x=-3,x= 3/2
Find the zeros of -2x^2-3x+9=0
The zeros of -2x^2-3x+9=0 are x=-3,x= 3/2