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

The solutions to the equation 2x^2-5x+2=0,x-2<0 are x=2,x= 1/2

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

The zeros of 2x^2-5x+2=0,x-2<0 are x=2,x= 1/2

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The solutions to the equation 2x^2-5x+2=0,x-2<0 are x=2,x= 1/2

The zeros of 2x^2-5x+2=0,x-2<0 are x=2,x= 1/2

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

x = 2, x = \frac{1}{2}

Decimal Notation

x = 2, x = 0.5

Solution steps

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 \cdot 2 = 3

x_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

x = \frac{- ( - 5 ) + 3}{2 \cdot 2} : 2

x = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

The solutions to the quadratic equation are:

x = 2, x = \frac{1}{2}

x, x^{2} + (q - 2) x + 10 - q = 0

x, f x = 3 x^{2} + 5 x - 2

x^{2} + 1 = \frac{41}{25}

5 \cdot x^{2} + 8 x - 45 = 0

t, m \cdot r u \cdot v s = 2 - 4 t + t^{2}

What are the solutions to the equation 2x^2-5x+2=0,x-2<0 ?
The solutions to the equation 2x^2-5x+2=0,x-2<0 are x=2,x= 1/2
Find the zeros of 2x^2-5x+2=0,x-2<0
The zeros of 2x^2-5x+2=0,x-2<0 are x=2,x= 1/2