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

The solutions to the equation x^2-4=2x are x=1+sqrt(5),x=1-sqrt(5)

Find the zeros of x^2-4=2x

The zeros of x^2-4=2x are x=1+sqrt(5),x=1-sqrt(5)

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x^{2} - 4 = 2 x

x = 1 + 5, x = 1 - 5

Decimal Notation

x = 3.23606 \dots, x = - 1.23606 \dots

Solution steps

x^{2} - 4 = 2 x

Move

2 x

to the left side

x^{2} - 4 - 2 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 2 x - 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

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

Separate the solutions

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

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

x = \frac{- ( - 2 ) - 2 5}{2 \cdot 1} : 1 - 5

The solutions to the quadratic equation are:

x = 1 + 5, x = 1 - 5

3 x + 4 = - 12

∣ x + 5 ∣ < 6

s^{2} - 81

6 x^{3} - 24 x y^{2}

8 (- 2) - 6 y = 24

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