What are the solutions to the equation 2x^2-9x+8=0 ?

The solutions to the equation 2x^2-9x+8=0 are x=(9+sqrt(17))/4 ,x=(9-sqrt(17))/4

Find the zeros of 2x^2-9x+8=0

The zeros of 2x^2-9x+8=0 are x=(9+sqrt(17))/4 ,x=(9-sqrt(17))/4

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2 x^{2} - 9 x + 8 = 0

x = \frac{9 + 17}{4}, x = \frac{9 - 17}{4}

Decimal Notation

x = 3.28077 \dots, x = 1.21922 \dots

Solution steps

2 x^{2} - 9 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 8 = 17

x_{1, 2} = \frac{- ( - 9 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 17}{2 \cdot 2}

x = \frac{- ( - 9 ) + 17}{2 \cdot 2} : \frac{9 + 17}{4}

x = \frac{- ( - 9 ) - 17}{2 \cdot 2} : \frac{9 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 17}{4}, x = \frac{9 - 17}{4}

7 a^{2} - 28 b^{2}

6 + 23 + 27 - 12

(x + 5)^{2} \leq 1 0^{2}

x^{2} - 16 x + 59 = 0

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

What are the solutions to the equation 2x^2-9x+8=0 ?
The solutions to the equation 2x^2-9x+8=0 are x=(9+sqrt(17))/4 ,x=(9-sqrt(17))/4
Find the zeros of 2x^2-9x+8=0
The zeros of 2x^2-9x+8=0 are x=(9+sqrt(17))/4 ,x=(9-sqrt(17))/4