What are the solutions to the equation 8x^2-65x+35=0 ?

The solutions to the equation 8x^2-65x+35=0 are x=(65+3sqrt(345))/(16),x=(65-3sqrt(345))/(16)

Find the zeros of 8x^2-65x+35=0

The zeros of 8x^2-65x+35=0 are x=(65+3sqrt(345))/(16),x=(65-3sqrt(345))/(16)

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8 x^{2} - 65 x + 35 = 0

x = \frac{65 + 3 345}{16}, x = \frac{65 - 3 345}{16}

Decimal Notation

x = 7.54515 \dots, x = 0.57984 \dots

Solution steps

8 x^{2} - 65 x + 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 65 ) \pm ( - 65 ) ^{2} - 4 \cdot 8 \cdot 35}{2 \cdot 8}

(- 65)^{2} - 4 \cdot 8 \cdot 35 = 3345

x_{1, 2} = \frac{- ( - 65 ) \pm 3 345}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 65 ) + 3 345}{2 \cdot 8}, x_{2} = \frac{- ( - 65 ) - 3 345}{2 \cdot 8}

x = \frac{- ( - 65 ) + 3 345}{2 \cdot 8} : \frac{65 + 3 345}{16}

x = \frac{- ( - 65 ) - 3 345}{2 \cdot 8} : \frac{65 - 3 345}{16}

The solutions to the quadratic equation are:

x = \frac{65 + 3 345}{16}, x = \frac{65 - 3 345}{16}

ln (\frac{1}{e ^{3}})

p + \frac{1}{3} (p + 1) + \frac{1}{4} (p + 2) = \frac{5}{6}

x^{2} - 6 x y + 8 y^{2}

\frac{a ^{- 5} b ^{- 4}}{a ^{- 13} b ^{8}}

c^{2} = 2 1^{2} + 2 1^{2}

What are the solutions to the equation 8x^2-65x+35=0 ?
The solutions to the equation 8x^2-65x+35=0 are x=(65+3sqrt(345))/(16),x=(65-3sqrt(345))/(16)
Find the zeros of 8x^2-65x+35=0
The zeros of 8x^2-65x+35=0 are x=(65+3sqrt(345))/(16),x=(65-3sqrt(345))/(16)