What are the solutions to the equation x^2-3x=8 ?

The solutions to the equation x^2-3x=8 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2

Find the zeros of x^2-3x=8

The zeros of x^2-3x=8 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2

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x^{2} - 3 x = 8

x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

Decimal Notation

x = 4.70156 \dots, x = - 1.70156 \dots

Solution steps

x^{2} - 3 x = 8

Move

8

to the left side

x^{2} - 3 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 8) = 41

x_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 1}

x = \frac{- ( - 3 ) + 41}{2 \cdot 1} : \frac{3 + 41}{2}

x = \frac{- ( - 3 ) - 41}{2 \cdot 1} : \frac{3 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

\frac{36 c ^{5} d ^{4}}{96 c d ^{8}}

\frac{2 x ^{2} - 1}{5 x - 2} \leq \frac{4 x + 1}{10}

\frac{a ^{- 16}}{a ^{- 4}}

x^{2} - 16 x + 63 > 0

\frac{x ^{4} y ^{3}}{x ^{2} y ^{3}}

What are the solutions to the equation x^2-3x=8 ?
The solutions to the equation x^2-3x=8 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2
Find the zeros of x^2-3x=8
The zeros of x^2-3x=8 are x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2