What are the solutions to the equation 2x^2+4x=x^2-3x-15 ?

The solutions to the equation 2x^2+4x=x^2-3x-15 are x=-7/2-i(sqrt(11))/2 ,x=-7/2+i(sqrt(11))/2

Find the zeros of 2x^2+4x=x^2-3x-15

The zeros of 2x^2+4x=x^2-3x-15 are x=-7/2-i(sqrt(11))/2 ,x=-7/2+i(sqrt(11))/2

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2 x^{2} + 4 x = x^{2} - 3 x - 15

x = - \frac{7}{2} - i \frac{11}{2}, x = - \frac{7}{2} + i \frac{11}{2}

Solution steps

2 x^{2} + 4 x = x^{2} - 3 x - 15

Switch sides

x^{2} - 3 x - 15 = 2 x^{2} + 4 x

Move

4 x

to the left side

x^{2} - 7 x - 15 = 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} - 7 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 1 ) ( - 15 )}{2 ( - 1 )}

Simplify

(- 7)^{2} - 4 (- 1) (- 15) : 11 i

x_{1, 2} = \frac{- ( - 7 ) \pm 11 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 11 i}{2 ( - 1 )}, x_{2} = \frac{- ( - 7 ) - 11 i}{2 ( - 1 )}

x = \frac{- ( - 7 ) + 11 i}{2 ( - 1 )} : - \frac{7}{2} - i \frac{11}{2}

x = \frac{- ( - 7 ) - 11 i}{2 ( - 1 )} : - \frac{7}{2} + i \frac{11}{2}

The solutions to the quadratic equation are:

x = - \frac{7}{2} - i \frac{11}{2}, x = - \frac{7}{2} + i \frac{11}{2}

\frac{y - 9}{2} < - 4

b^{2} - 5 b - 14 = 0

(x + 3)^{2} = 25

3 x^{2} - 2 x + 7 = 0

k - 7 = - 12

What are the solutions to the equation 2x^2+4x=x^2-3x-15 ?
The solutions to the equation 2x^2+4x=x^2-3x-15 are x=-7/2-i(sqrt(11))/2 ,x=-7/2+i(sqrt(11))/2
Find the zeros of 2x^2+4x=x^2-3x-15
The zeros of 2x^2+4x=x^2-3x-15 are x=-7/2-i(sqrt(11))/2 ,x=-7/2+i(sqrt(11))/2