What are the solutions to the equation 7=-x^2-4x ?

The solutions to the equation 7=-x^2-4x are x=-2-sqrt(3)i,x=-2+sqrt(3)i

Find the zeros of 7=-x^2-4x

The zeros of 7=-x^2-4x are x=-2-sqrt(3)i,x=-2+sqrt(3)i

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7 = - x^{2} - 4 x

x = - 2 - 3 i, x = - 2 + 3 i

Solution steps

7 = - x^{2} - 4 x

Switch sides

- x^{2} - 4 x = 7

Move

7

to the left side

- x^{2} - 4 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) ( - 7 )}{2 ( - 1 )}

Simplify

(- 4)^{2} - 4 (- 1) (- 7) : 23 i

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 3 i}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 2 3 i}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

x = - 2 - 3 i, x = - 2 + 3 i

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

2 x^{2} - 36 x + 64 = 0

t, s = u t + \frac{1}{2} a t^{2}

3 x^{2} - 4 x + 6 = - 9 x + 4

- 2 x^{2} + 3 x - 1 = 0

What are the solutions to the equation 7=-x^2-4x ?
The solutions to the equation 7=-x^2-4x are x=-2-sqrt(3)i,x=-2+sqrt(3)i
Find the zeros of 7=-x^2-4x
The zeros of 7=-x^2-4x are x=-2-sqrt(3)i,x=-2+sqrt(3)i