What are the solutions to the equation -12+57=-5x(x+2) ?

The solutions to the equation -12+57=-5x(x+2) are x=-1-2sqrt(2)i,x=-1+2sqrt(2)i

Find the zeros of -12+57=-5x(x+2)

The zeros of -12+57=-5x(x+2) are x=-1-2sqrt(2)i,x=-1+2sqrt(2)i

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- 12 + 57 = - 5 x (x + 2)

x = - 1 - 22 i, x = - 1 + 22 i

Solution steps

- 12 + 57 = - 5 x (x + 2)

Expand

- 5 x (x + 2) : - 5 x^{2} - 10 x

45 = - 5 x^{2} - 10 x

Switch sides

- 5 x^{2} - 10 x = 45

Move

45

to the left side

- 5 x^{2} - 10 x - 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 5 ) ( - 45 )}{2 ( - 5 )}

Simplify

(- 10)^{2} - 4 (- 5) (- 45) : 202 i

x_{1, 2} = \frac{- ( - 10 ) \pm 20 2 i}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 20 2 i}{2 ( - 5 )}, x_{2} = \frac{- ( - 10 ) - 20 2 i}{2 ( - 5 )}

x = \frac{- ( - 10 ) + 20 2 i}{2 ( - 5 )} : - 1 - 22 i

x = \frac{- ( - 10 ) - 20 2 i}{2 ( - 5 )} : - 1 + 22 i

The solutions to the quadratic equation are:

x = - 1 - 22 i, x = - 1 + 22 i

(3 x + 3)^{2} = 25

x, x^{2} = 4 x y - 1

x, x^{2} = 4 x y - 3

4 x^{2} = 32 x - 55

(t - 5)^{2} - \frac{1}{4} = 0

What are the solutions to the equation -12+57=-5x(x+2) ?
The solutions to the equation -12+57=-5x(x+2) are x=-1-2sqrt(2)i,x=-1+2sqrt(2)i
Find the zeros of -12+57=-5x(x+2)
The zeros of -12+57=-5x(x+2) are x=-1-2sqrt(2)i,x=-1+2sqrt(2)i