What are the solutions to the equation 2x^2-13x+17=0 ?

The solutions to the equation 2x^2-13x+17=0 are x=(13+sqrt(33))/4 ,x=(13-sqrt(33))/4

Find the zeros of 2x^2-13x+17=0

The zeros of 2x^2-13x+17=0 are x=(13+sqrt(33))/4 ,x=(13-sqrt(33))/4

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2 x^{2} - 13 x + 17 = 0

x = \frac{13 + 33}{4}, x = \frac{13 - 33}{4}

Decimal Notation

x = 4.68614 \dots, x = 1.81385 \dots

Solution steps

2 x^{2} - 13 x + 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 17}{2 \cdot 2}

(- 13)^{2} - 4 \cdot 2 \cdot 17 = 33

x_{1, 2} = \frac{- ( - 13 ) \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 33}{2 \cdot 2}, x_{2} = \frac{- ( - 13 ) - 33}{2 \cdot 2}

x = \frac{- ( - 13 ) + 33}{2 \cdot 2} : \frac{13 + 33}{4}

x = \frac{- ( - 13 ) - 33}{2 \cdot 2} : \frac{13 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{13 + 33}{4}, x = \frac{13 - 33}{4}

\frac{1}{2} (2)^{- 2}

5 x - 3 > 7

5 t^{2} - 40 t = - 15

2 k^{2} - 16 k^{2} - 3 k + 24

\frac{1}{4} (2)

What are the solutions to the equation 2x^2-13x+17=0 ?
The solutions to the equation 2x^2-13x+17=0 are x=(13+sqrt(33))/4 ,x=(13-sqrt(33))/4
Find the zeros of 2x^2-13x+17=0
The zeros of 2x^2-13x+17=0 are x=(13+sqrt(33))/4 ,x=(13-sqrt(33))/4