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

The solutions to the equation 2x+3=-x^2 are x=-1-sqrt(2)i,x=-1+sqrt(2)i

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

The zeros of 2x+3=-x^2 are x=-1-sqrt(2)i,x=-1+sqrt(2)i

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2 x + 3 = - x^{2}

x = - 1 - 2 i, x = - 1 + 2 i

Solution steps

2 x + 3 = - x^{2}

Switch sides

- x^{2} = 2 x + 3

Move

3

to the left side

- x^{2} - 3 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 (- 1) (- 3) : 22 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 - 2 i, x = - 1 + 2 i

0.5 (x - 1)^{2} - 0.25 (x + 1)^{2} = 4 - x

t, s = - 16 t^{2} + 108 t

x^{2} + 28 x - 128 = 0

2 x^{2} + 5 x - 58 = 0

- 8 x^{2} + 3 x^{2} + 3 x - 1 = - 9 x - 8 - 6 x - 7 x^{2} + 7

What are the solutions to the equation 2x+3=-x^2 ?
The solutions to the equation 2x+3=-x^2 are x=-1-sqrt(2)i,x=-1+sqrt(2)i
Find the zeros of 2x+3=-x^2
The zeros of 2x+3=-x^2 are x=-1-sqrt(2)i,x=-1+sqrt(2)i