What are the solutions to the equation x^2-4x-7x+(x-1)^2=7+8-3x ?

The solutions to the equation x^2-4x-7x+(x-1)^2=7+8-3x are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2

Find the zeros of x^2-4x-7x+(x-1)^2=7+8-3x

The zeros of x^2-4x-7x+(x-1)^2=7+8-3x are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2

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x^{2} - 4 x - 7 x + (x - 1)^{2} = 7 + 8 - 3 x

x = \frac{5 + 53}{2}, x = \frac{5 - 53}{2}

Decimal Notation

x = 6.14005 \dots, x = - 1.14005 \dots

Solution steps

x^{2} - 4 x - 7 x + (x - 1)^{2} = 7 + 8 - 3 x

Expand

x^{2} - 4 x - 7 x + (x - 1)^{2} : 2 x^{2} - 13 x + 1

2 x^{2} - 13 x + 1 = - 3 x + 15

Move

15

to the left side

2 x^{2} - 13 x - 14 = - 3 x

Move

3 x

to the left side

2 x^{2} - 10 x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 ( - 14 )}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 (- 14) = 253

x_{1, 2} = \frac{- ( - 10 ) \pm 2 53}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 53}{2 \cdot 2}, x_{2} = \frac{- ( - 10 ) - 2 53}{2 \cdot 2}

x = \frac{- ( - 10 ) + 2 53}{2 \cdot 2} : \frac{5 + 53}{2}

x = \frac{- ( - 10 ) - 2 53}{2 \cdot 2} : \frac{5 - 53}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 53}{2}, x = \frac{5 - 53}{2}

2 n^{2} + 6 n - 4 = 0

x, y^{3} - 9 y^{2} + 15 y = x^{2} + 10 x - 8

(x - 6)^{2} = 5

x^{2} - 1 - 1 = 0

r, y = 2 (x + r)^{2} - s

What are the solutions to the equation x^2-4x-7x+(x-1)^2=7+8-3x ?
The solutions to the equation x^2-4x-7x+(x-1)^2=7+8-3x are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2
Find the zeros of x^2-4x-7x+(x-1)^2=7+8-3x
The zeros of x^2-4x-7x+(x-1)^2=7+8-3x are x=(5+sqrt(53))/2 ,x=(5-sqrt(53))/2