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

The solutions to the equation quadraticformula x^2-5x+2=0 are x=(5+sqrt(17))/2 ,x=(5-sqrt(17))/2

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

The zeros of quadraticformula x^2-5x+2=0 are x=(5+sqrt(17))/2 ,x=(5-sqrt(17))/2

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

x = \frac{5 + 17}{2}, x = \frac{5 - 17}{2}

Decimal Notation

x = 4.56155 \dots, x = 0.43844 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 1 \cdot 2 = 17

x_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 1}

x = \frac{- ( - 5 ) + 17}{2 \cdot 1} : \frac{5 + 17}{2}

x = \frac{- ( - 5 ) - 17}{2 \cdot 1} : \frac{5 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 17}{2}, x = \frac{5 - 17}{2}

x^{2} + 18 x = 7

x^{2} - 14 x - 12 = 4

m^{2} + 12 = 0

x^{2} + 3 x + 8

x, f = 4 x^{2} - 4 x y + y^{2}

What are the solutions to the equation quadraticformula x^2-5x+2=0 ?
The solutions to the equation quadraticformula x^2-5x+2=0 are x=(5+sqrt(17))/2 ,x=(5-sqrt(17))/2
Find the zeros of quadraticformula x^2-5x+2=0
The zeros of quadraticformula x^2-5x+2=0 are x=(5+sqrt(17))/2 ,x=(5-sqrt(17))/2