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

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

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

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

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x^{2} - 4 x = - 3

x = 3, x = 1

Solution steps

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

- 3 = - \frac{3}{2} x + 3

- \frac{4}{3 ^{2}}

\frac{n}{19} = 7

8 a^{3} - 27

x^{2} + (\frac{7}{25})^{2} = 1

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