What are the solutions to the equation (2x-3)^2-14=2x(x-7) ?

The solutions to the equation (2x-3)^2-14=2x(x-7) are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2

Find the zeros of (2x-3)^2-14=2x(x-7)

The zeros of (2x-3)^2-14=2x(x-7) are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2

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(2 x - 3)^{2} - 14 = 2 x (x - 7)

x = \frac{- 1 + 11}{2}, x = - \frac{1 + 11}{2}

Decimal Notation

x = 1.15831 \dots, x = - 2.15831 \dots

Solution steps

(2 x - 3)^{2} - 14 = 2 x (x - 7)

Expand

(2 x - 3)^{2} - 14 : 4 x^{2} - 12 x - 5

Expand

2 x (x - 7) : 2 x^{2} - 14 x

4 x^{2} - 12 x - 5 = 2 x^{2} - 14 x

Move

14 x

to the left side

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

Move

2 x^{2}

to the left side

2 x^{2} + 2 x - 5 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 5) = 211

x_{1, 2} = \frac{- 2 \pm 2 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 11}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 11}{2 \cdot 2}

x = \frac{- 2 + 2 11}{2 \cdot 2} : \frac{- 1 + 11}{2}

x = \frac{- 2 - 2 11}{2 \cdot 2} : - \frac{1 + 11}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 11}{2}, x = - \frac{1 + 11}{2}

3 x^{2} - 30 x + 63 = 0

x^{2} - 18 x = - 66

x^{2} + \frac{1}{2} x + 3

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

4 (x^{2} - 4) - 5 (3 + 2 x) = - 7

What are the solutions to the equation (2x-3)^2-14=2x(x-7) ?
The solutions to the equation (2x-3)^2-14=2x(x-7) are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2
Find the zeros of (2x-3)^2-14=2x(x-7)
The zeros of (2x-3)^2-14=2x(x-7) are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2