What are the solutions to the equation 1=x^2-3x ?

The solutions to the equation 1=x^2-3x are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2

Find the zeros of 1=x^2-3x

The zeros of 1=x^2-3x are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2

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1 = x^{2} - 3 x

x = \frac{3 + 13}{2}, x = \frac{3 - 13}{2}

Decimal Notation

x = 3.30277 \dots, x = - 0.30277 \dots

Solution steps

1 = x^{2} - 3 x

Switch sides

x^{2} - 3 x = 1

Move

1

to the left side

x^{2} - 3 x - 1 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 1) = 13

x_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

x = \frac{- ( - 3 ) + 13}{2 \cdot 1} : \frac{3 + 13}{2}

x = \frac{- ( - 3 ) - 13}{2 \cdot 1} : \frac{3 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 13}{2}, x = \frac{3 - 13}{2}

4 y = 24

64 - 81 u^{2}

- 2 x - 8 < 2 x + 4

0 \leq \frac{4 - x}{3} \leq 2

\frac{4 b}{15 x ^{3} y ^{2}} - \frac{3 b}{35 x ^{2} y ^{4} z}

What are the solutions to the equation 1=x^2-3x ?
The solutions to the equation 1=x^2-3x are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2
Find the zeros of 1=x^2-3x
The zeros of 1=x^2-3x are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2