What are the solutions to the equation 4=-x^2+x+6 ?

The solutions to the equation 4=-x^2+x+6 are x=-1,x=2

Find the zeros of 4=-x^2+x+6

The zeros of 4=-x^2+x+6 are x=-1,x=2

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

x = - 1, x = 2

Solution steps

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

Switch sides

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

Move

4

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = 2

x, x^{2} + y^{2} + z^{2} = a^{2}

x^{2} - 9 x - 136 = 0

x^{2} - \frac{1}{2} x - 2 = 0

x, x y^{2} - 2 x^{2} y = 4

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

What are the solutions to the equation 4=-x^2+x+6 ?
The solutions to the equation 4=-x^2+x+6 are x=-1,x=2
Find the zeros of 4=-x^2+x+6
The zeros of 4=-x^2+x+6 are x=-1,x=2