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

The solutions to the equation 2x^2-13x+2=0 are x=(13+3sqrt(17))/4 ,x=(13-3sqrt(17))/4

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

The zeros of 2x^2-13x+2=0 are x=(13+3sqrt(17))/4 ,x=(13-3sqrt(17))/4

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

x = \frac{13 + 3 17}{4}, x = \frac{13 - 3 17}{4}

Decimal Notation

x = 6.34232 \dots, x = 0.15767 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 13)^{2} - 4 \cdot 2 \cdot 2 = 317

x_{1, 2} = \frac{- ( - 13 ) \pm 3 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 3 17}{2 \cdot 2}, x_{2} = \frac{- ( - 13 ) - 3 17}{2 \cdot 2}

x = \frac{- ( - 13 ) + 3 17}{2 \cdot 2} : \frac{13 + 3 17}{4}

x = \frac{- ( - 13 ) - 3 17}{2 \cdot 2} : \frac{13 - 3 17}{4}

The solutions to the quadratic equation are:

x = \frac{13 + 3 17}{4}, x = \frac{13 - 3 17}{4}

x^{2} - 10 x = 0

3 x^{2} - 4 x = 0

y, x^{2} + y^{2} = 16

(x + 4) (x + 2) = 15

4 w^{2} + 4 w + 1 = 121

What are the solutions to the equation 2x^2-13x+2=0 ?
The solutions to the equation 2x^2-13x+2=0 are x=(13+3sqrt(17))/4 ,x=(13-3sqrt(17))/4
Find the zeros of 2x^2-13x+2=0
The zeros of 2x^2-13x+2=0 are x=(13+3sqrt(17))/4 ,x=(13-3sqrt(17))/4