What are the solutions to the equation 2x^2-5x=x^2+2x-1 ?

The solutions to the equation 2x^2-5x=x^2+2x-1 are x=(7-3sqrt(5))/2 ,x=(7+3sqrt(5))/2

Find the zeros of 2x^2-5x=x^2+2x-1

The zeros of 2x^2-5x=x^2+2x-1 are x=(7-3sqrt(5))/2 ,x=(7+3sqrt(5))/2

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2 x^{2} - 5 x = x^{2} + 2 x - 1

x = \frac{7 - 3 5}{2}, x = \frac{7 + 3 5}{2}

Decimal

x = 0.14589 \dots, x = 6.85410 \dots

Solution steps

2 x^{2} - 5 x = x^{2} + 2 x - 1

Switch sides

x^{2} + 2 x - 1 = 2 x^{2} - 5 x

Move

5 x

to the left side

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

Move

2 x^{2}

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 ) ( - 1 )}{2 ( - 1 )}

7^{2} - 4 (- 1) (- 1) = 35

x_{1, 2} = \frac{- 7 \pm 3 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 7 + 3 5}{2 ( - 1 )}, x_{2} = \frac{- 7 - 3 5}{2 ( - 1 )}

x = \frac{- 7 + 3 5}{2 ( - 1 )} : \frac{7 - 3 5}{2}

x = \frac{- 7 - 3 5}{2 ( - 1 )} : \frac{7 + 3 5}{2}

The solutions to the quadratic equation are:

x = \frac{7 - 3 5}{2}, x = \frac{7 + 3 5}{2}

9 x^{2} - 12 x = 14

(x - 2) (2 x + 4) = 12

2 x^{2} - 1 = 11 x + 5

3 x^{2} - 12 x + 13

(100 - 2 x) (50 - 2 x) = 3600

What are the solutions to the equation 2x^2-5x=x^2+2x-1 ?
The solutions to the equation 2x^2-5x=x^2+2x-1 are x=(7-3sqrt(5))/2 ,x=(7+3sqrt(5))/2
Find the zeros of 2x^2-5x=x^2+2x-1
The zeros of 2x^2-5x=x^2+2x-1 are x=(7-3sqrt(5))/2 ,x=(7+3sqrt(5))/2