What are the solutions to the equation 2x^2-8x=-2 ?

The solutions to the equation 2x^2-8x=-2 are x=2+sqrt(3),x=2-sqrt(3)

Find the zeros of 2x^2-8x=-2

The zeros of 2x^2-8x=-2 are x=2+sqrt(3),x=2-sqrt(3)

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2 x^{2} - 8 x = - 2

x = 2 + 3, x = 2 - 3

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

2 x^{2} - 8 x = - 2

Move

2

to the left side

2 x^{2} - 8 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 2 = 43

x_{1, 2} = \frac{- ( - 8 ) \pm 4 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 3}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 4 3}{2 \cdot 2}

x = \frac{- ( - 8 ) + 4 3}{2 \cdot 2} : 2 + 3

x = \frac{- ( - 8 ) - 4 3}{2 \cdot 2} : 2 - 3

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

x^{2} - x + 6 = 3

12 x^{2} - 120 x + 216 = 0

- 7 x - 10 = 3 x^{2}

x^{2} - 1.6 x + 0.63 = 0

\frac{4 ( t - 1.4 ) ^{2}}{2} = \frac{87 t}{3.6}

What are the solutions to the equation 2x^2-8x=-2 ?
The solutions to the equation 2x^2-8x=-2 are x=2+sqrt(3),x=2-sqrt(3)
Find the zeros of 2x^2-8x=-2
The zeros of 2x^2-8x=-2 are x=2+sqrt(3),x=2-sqrt(3)