What are the solutions to the equation 2x^2-8x+5=0 ?

The solutions to the equation 2x^2-8x+5=0 are x=(4+sqrt(6))/2 ,x=(4-sqrt(6))/2

Find the zeros of 2x^2-8x+5=0

The zeros of 2x^2-8x+5=0 are x=(4+sqrt(6))/2 ,x=(4-sqrt(6))/2

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2 x^{2} - 8 x + 5 = 0

x = \frac{4 + 6}{2}, x = \frac{4 - 6}{2}

Decimal Notation

x = 3.22474 \dots, x = 0.77525 \dots

Solution steps

2 x^{2} - 8 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 5 = 26

x_{1, 2} = \frac{- ( - 8 ) \pm 2 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 6}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 2 6}{2 \cdot 2}

x = \frac{- ( - 8 ) + 2 6}{2 \cdot 2} : \frac{4 + 6}{2}

x = \frac{- ( - 8 ) - 2 6}{2 \cdot 2} : \frac{4 - 6}{2}

The solutions to the quadratic equation are:

x = \frac{4 + 6}{2}, x = \frac{4 - 6}{2}

- 12 x^{4} + 32 x^{3} - 15 x^{2} - 3 x + 2

\frac{3}{5} x - \frac{7}{5} x + \frac{5}{3} = \frac{31}{15}

(- \frac{3}{5})^{3}

- \frac{5}{2} + 5

y^{2} - y - 42

What are the solutions to the equation 2x^2-8x+5=0 ?
The solutions to the equation 2x^2-8x+5=0 are x=(4+sqrt(6))/2 ,x=(4-sqrt(6))/2
Find the zeros of 2x^2-8x+5=0
The zeros of 2x^2-8x+5=0 are x=(4+sqrt(6))/2 ,x=(4-sqrt(6))/2