What are the solutions to the equation x^2-3x+2=0,(1,-1,2,-2) ?

The solutions to the equation x^2-3x+2=0,(1,-1,2,-2) are x=2,x=1

Find the zeros of x^2-3x+2=0,(1,-1,2,-2)

The zeros of x^2-3x+2=0,(1,-1,2,-2) are x=2,x=1

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x^{2} - 3 x + 2 = 0, (1, - 1, 2, - 2)

x = 2, x = 1

Solution steps

x^{2} - 3 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = 1

2 m^{2} + 8 m + 8 = 0

- 5 t^{2} + 9 t + 3 = 0

2 x^{2} + 20 x - 8 = 0

321.0069 x^{2} - 5130.1778 x + 14682.6031 = 0

x (5 x - 2) = 24

What are the solutions to the equation x^2-3x+2=0,(1,-1,2,-2) ?
The solutions to the equation x^2-3x+2=0,(1,-1,2,-2) are x=2,x=1
Find the zeros of x^2-3x+2=0,(1,-1,2,-2)
The zeros of x^2-3x+2=0,(1,-1,2,-2) are x=2,x=1