What are the solutions to the equation -3=x^2-4x ?

The solutions to the equation -3=x^2-4x are x=3,x=1

Find the zeros of -3=x^2-4x

The zeros of -3=x^2-4x are x=3,x=1

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- 3 = x^{2} - 4 x

x = 3, x = 1

Solution steps

- 3 = x^{2} - 4 x

Switch sides

x^{2} - 4 x = - 3

Move

3

to the left side

x^{2} - 4 x + 3 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

x = \frac{- ( - 4 ) - 2}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

\frac{a}{c} (3 b)

\frac{x ^{5} y ^{6}}{x y ^{2}}

(\frac{- 1 + 3 i}{2})^{3}

7 = \frac{1}{3} (8 x + 3)

(- 1)^{\frac{1}{2}}

What are the solutions to the equation -3=x^2-4x ?
The solutions to the equation -3=x^2-4x are x=3,x=1
Find the zeros of -3=x^2-4x
The zeros of -3=x^2-4x are x=3,x=1