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

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

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

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

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

x = 2, x = 1

Solution steps

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

Refine

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

5 t^{2} - 5 t - 30 = 0

(x + 1) (x - 3) = 2 (x + 5) - 16

9 m^{2} = 5 m^{2} - 10

3 x^{2} - 7 x - 12 = 0

(x - 5) (x + 2) = - 6

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