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

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

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

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

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x^{2} - 4 x = - 2

x = 2 + 2, x = 2 - 2

Decimal

x = 3.41421 \dots, x = 0.58578 \dots

Solution steps

x^{2} - 4 x = - 2

Move

2

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 2 = 22

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 2}{2 \cdot 1} : 2 + 2

x = \frac{- ( - 4 ) - 2 2}{2 \cdot 1} : 2 - 2

The solutions to the quadratic equation are:

x = 2 + 2, x = 2 - 2

b, \frac{1}{4} = b^{- 2}

75 x^{10}

- 10/0.01

\frac{1}{2} \cdot 2.5

\frac{( 2 π )}{4}

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