What are the solutions to the equation -2=x^2-3x ?

The solutions to the equation -2=x^2-3x are x=2,x=1

Find the zeros of -2=x^2-3x

The zeros of -2=x^2-3x are x=2,x=1

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- 2 = x^{2} - 3 x

x = 2, x = 1

Solution steps

- 2 = x^{2} - 3 x

Switch sides

x^{2} - 3 x = - 2

Move

2

to the left side

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 x^{2} + x - 136 = 0

x + \frac{x ^{2} + 4}{5} = 2

x^{2} + 22 x - 4 = 0

- 25 x^{2} + 4 x = 0

a, 14 (a^{2} + b^{2} + c^{2}) = (a + 2 b + 3 c)^{2}

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