What are the solutions to the equation -2x^2-3x+20=0 ?

The solutions to the equation -2x^2-3x+20=0 are x=-4,x= 5/2

Find the zeros of -2x^2-3x+20=0

The zeros of -2x^2-3x+20=0 are x=-4,x= 5/2

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- 2 x^{2} - 3 x + 20 = 0

x = - 4, x = \frac{5}{2}

Decimal Notation

x = - 4, x = 2.5

Solution steps

- 2 x^{2} - 3 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 20}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 20 = 13

x_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 13}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 13}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 13}{2 ( - 2 )} : - 4

x = \frac{- ( - 3 ) - 13}{2 ( - 2 )} : \frac{5}{2}

The solutions to the quadratic equation are:

x = - 4, x = \frac{5}{2}

x^{2} - 2 (x - 6) = 4 x + 4

6 (5 x + x)^{2} - 42 = 0

7 x = 3 x^{2}

a^{2} = 225

2 x^{2} + 6 x - 8

What are the solutions to the equation -2x^2-3x+20=0 ?
The solutions to the equation -2x^2-3x+20=0 are x=-4,x= 5/2
Find the zeros of -2x^2-3x+20=0
The zeros of -2x^2-3x+20=0 are x=-4,x= 5/2