What are the solutions to the equation -x^2-3x+4=0 ?

The solutions to the equation -x^2-3x+4=0 are x=-4,x=1

Find the zeros of -x^2-3x+4=0

The zeros of -x^2-3x+4=0 are x=-4,x=1

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- x^{2} - 3 x + 4 = 0

x = - 4, x = 1

Solution steps

- x^{2} - 3 x + 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 1) \cdot 4 = 5

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

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 ( - 1 )}, x_{2} = \frac{- ( - 3 ) - 5}{2 ( - 1 )}

x = \frac{- ( - 3 ) + 5}{2 ( - 1 )} : - 4

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

The solutions to the quadratic equation are:

x = - 4, x = 1

3 < 4 s - 1

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

a^{2} + ab + 2 a + 2 b

81 x = 1984.87

∣ 7 x - 1 ∣ < 0

What are the solutions to the equation -x^2-3x+4=0 ?
The solutions to the equation -x^2-3x+4=0 are x=-4,x=1
Find the zeros of -x^2-3x+4=0
The zeros of -x^2-3x+4=0 are x=-4,x=1