What are the solutions to the equation solvefor x,-x^2+3x+40=0 ?

The solutions to the equation solvefor x,-x^2+3x+40=0 are x=-5,x=8

Find the zeros of solvefor x,-x^2+3x+40=0

The zeros of solvefor x,-x^2+3x+40=0 are x=-5,x=8

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x, - x^{2} + 3 x + 40 = 0

x = - 5, x = 8

Solution steps

- x^{2} + 3 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) \cdot 40}{2 ( - 1 )}

3^{2} - 4 (- 1) \cdot 40 = 13

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

Separate the solutions

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

x = \frac{- 3 + 13}{2 ( - 1 )} : - 5

x = \frac{- 3 - 13}{2 ( - 1 )} : 8

The solutions to the quadratic equation are:

x = - 5, x = 8

0 = - 16 t^{2} + 96 t + 112

4 x^{2} + 16 x - 5 = 0

x^{2} - 16 x + 17 = 0

x^{2} + 5 x - 21 = 0

x, 2088.4 x^{2} - 15070 x + 5697.2 = 0

What are the solutions to the equation solvefor x,-x^2+3x+40=0 ?
The solutions to the equation solvefor x,-x^2+3x+40=0 are x=-5,x=8
Find the zeros of solvefor x,-x^2+3x+40=0
The zeros of solvefor x,-x^2+3x+40=0 are x=-5,x=8