What are the solutions to the equation x^2-x-48=0 ?

The solutions to the equation x^2-x-48=0 are x=(1+sqrt(193))/2 ,x=(1-sqrt(193))/2

Find the zeros of x^2-x-48=0

The zeros of x^2-x-48=0 are x=(1+sqrt(193))/2 ,x=(1-sqrt(193))/2

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x^{2} - x - 48 = 0

x = \frac{1 + 193}{2}, x = \frac{1 - 193}{2}

Decimal Notation

x = 7.44622 \dots, x = - 6.44622 \dots

Solution steps

x^{2} - x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 48) = 193

x_{1, 2} = \frac{- ( - 1 ) \pm 193}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 193}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 193}{2 \cdot 1}

x = \frac{- ( - 1 ) + 193}{2 \cdot 1} : \frac{1 + 193}{2}

x = \frac{- ( - 1 ) - 193}{2 \cdot 1} : \frac{1 - 193}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 193}{2}, x = \frac{1 - 193}{2}

- 0.01 x^{2} - 0.2 x + 8 = 0

x, y = - 0.00001 x^{2} + 0.01 x + 0.2001

2 m^{2} - 13 m - 45 = 0

3 a^{2} + 9 a - 30 = 0

x, 5 x^{2} + 3 = 2 y^{3} + 5

What are the solutions to the equation x^2-x-48=0 ?
The solutions to the equation x^2-x-48=0 are x=(1+sqrt(193))/2 ,x=(1-sqrt(193))/2
Find the zeros of x^2-x-48=0
The zeros of x^2-x-48=0 are x=(1+sqrt(193))/2 ,x=(1-sqrt(193))/2