What are the solutions to the equation -5+2x^2=-7x+x^2-4 ?

The solutions to the equation -5+2x^2=-7x+x^2-4 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2

Find the zeros of -5+2x^2=-7x+x^2-4

The zeros of -5+2x^2=-7x+x^2-4 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2

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- 5 + 2 x^{2} = - 7 x + x^{2} - 4

x = - \frac{7 + 53}{2}, x = \frac{53 - 7}{2}

Solution steps

- 5 + 2 x^{2} = - 7 x + x^{2} - 4

Switch sides

- 7 x + x^{2} - 4 = - 5 + 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} - 7 x - 4 = - 5

Move

5

to the left side

- x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 (- 1) \cdot 1 = 53

x_{1, 2} = \frac{- ( - 7 ) \pm 53}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 53}{2 ( - 1 )}, x_{2} = \frac{- ( - 7 ) - 53}{2 ( - 1 )}

x = \frac{- ( - 7 ) + 53}{2 ( - 1 )} : - \frac{7 + 53}{2}

x = \frac{- ( - 7 ) - 53}{2 ( - 1 )} : \frac{53 - 7}{2}

The solutions to the quadratic equation are:

x = - \frac{7 + 53}{2}, x = \frac{53 - 7}{2}

5 w^{2} + 10 w = 0

- 3 x^{2} - 9 = 0

(x - 3) (x + 2) + 9 x = 3 (x^{2} - 5) - 1

64 x^{2} + 144 x + 81 = 0

A = x^{2} + 4 x + 11

What are the solutions to the equation -5+2x^2=-7x+x^2-4 ?
The solutions to the equation -5+2x^2=-7x+x^2-4 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2
Find the zeros of -5+2x^2=-7x+x^2-4
The zeros of -5+2x^2=-7x+x^2-4 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2