What are the solutions to the equation 2x^2-8x=-7 ?

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

Find the zeros of 2x^2-8x=-7

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

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2 x^{2} - 8 x = - 7

x = \frac{4 + 2}{2}, x = \frac{4 - 2}{2}

Decimal Notation

x = 2.70710 \dots, x = 1.29289 \dots

Solution steps

2 x^{2} - 8 x = - 7

Move

7

to the left side

2 x^{2} - 8 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 7 = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{4 + 2}{2}, x = \frac{4 - 2}{2}

0 = - x^{2} + 3 x + 1

\frac{1}{12.5} + \frac{1}{45.5} + \frac{1}{85.5}

\frac{k ^{\frac{1}{3}}}{k ^{\frac{2}{3}} \cdot k ^{- 1}}

6 5^{2}

\frac{y ^{3}}{y ^{2}}

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