What are the solutions to the equation (x-1)^2-x(2x-5)=0 ?

The solutions to the equation (x-1)^2-x(2x-5)=0 are x=-(-3+sqrt(13))/2 ,x=(3+sqrt(13))/2

Find the zeros of (x-1)^2-x(2x-5)=0

The zeros of (x-1)^2-x(2x-5)=0 are x=-(-3+sqrt(13))/2 ,x=(3+sqrt(13))/2

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(x - 1)^{2} - x (2 x - 5) = 0

x = - \frac{- 3 + 13}{2}, x = \frac{3 + 13}{2}

Decimal Notation

x = - 0.30277 \dots, x = 3.30277 \dots

Solution steps

(x - 1)^{2} - x (2 x - 5) = 0

Expand

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

- x^{2} + 3 x + 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 (- 1) \cdot 1 = 13

x_{1, 2} = \frac{- 3 \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 3 + 13}{2 ( - 1 )}, x_{2} = \frac{- 3 - 13}{2 ( - 1 )}

x = \frac{- 3 + 13}{2 ( - 1 )} : - \frac{- 3 + 13}{2}

x = \frac{- 3 - 13}{2 ( - 1 )} : \frac{3 + 13}{2}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 13}{2}, x = \frac{3 + 13}{2}

x^{2} - 15 x - 16 = - 9 x - 9

36 x^{2} + 4 = 40

3 x^{2} + 8 = 13

x, f (x + 3) = x^{2} - 4 x + 4

x^{2} + 23 x - 140 = 0

What are the solutions to the equation (x-1)^2-x(2x-5)=0 ?
The solutions to the equation (x-1)^2-x(2x-5)=0 are x=-(-3+sqrt(13))/2 ,x=(3+sqrt(13))/2
Find the zeros of (x-1)^2-x(2x-5)=0
The zeros of (x-1)^2-x(2x-5)=0 are x=-(-3+sqrt(13))/2 ,x=(3+sqrt(13))/2