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

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

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

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

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

x = - 8, x = 5

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 )} : - 8

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

The solutions to the quadratic equation are:

x = - 8, x = 5

\frac{18}{9 i}

\frac{5}{x - 1} > - 3

ln (3.5)

\frac{9 ^{5}}{9}

2 a^{3} + 8 a^{2} - 64 a

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