What are the solutions to the equation x^2-16x-48=0 ?

The solutions to the equation x^2-16x-48=0 are x=4(2+sqrt(7)),x=4(2-sqrt(7))

Find the zeros of x^2-16x-48=0

The zeros of x^2-16x-48=0 are x=4(2+sqrt(7)),x=4(2-sqrt(7))

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x^{2} - 16 x - 48 = 0

x = 4 (2 + 7), x = 4 (2 - 7)

Decimal Notation

x = 18.58300 \dots, x = - 2.58300 \dots

Solution steps

x^{2} - 16 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot (- 48) = 87

x_{1, 2} = \frac{- ( - 16 ) \pm 8 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8 7}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 8 7}{2 \cdot 1}

x = \frac{- ( - 16 ) + 8 7}{2 \cdot 1} : 4 (2 + 7)

x = \frac{- ( - 16 ) - 8 7}{2 \cdot 1} : 4 (2 - 7)

The solutions to the quadratic equation are:

x = 4 (2 + 7), x = 4 (2 - 7)

x^{2} + 3 x - 5 = 4 x + 3

3 x^{2} - 6 x - 2 = 0

64 = r^{2}

4 x^{2} + 9 = x^{2} + 60

4 u^{2} - 8 u + 3 = 0

What are the solutions to the equation x^2-16x-48=0 ?
The solutions to the equation x^2-16x-48=0 are x=4(2+sqrt(7)),x=4(2-sqrt(7))
Find the zeros of x^2-16x-48=0
The zeros of x^2-16x-48=0 are x=4(2+sqrt(7)),x=4(2-sqrt(7))