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

The solutions to the equation 0=x^2+2x-48 are x=6,x=-8

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

The zeros of 0=x^2+2x-48 are x=6,x=-8

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

x = 6, x = - 8

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 48) = 14

x_{1, 2} = \frac{- 2 \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 14}{2 \cdot 1}, x_{2} = \frac{- 2 - 14}{2 \cdot 1}

x = \frac{- 2 + 14}{2 \cdot 1} : 6

x = \frac{- 2 - 14}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 6, x = - 8

3 a^{2} + 2 a = 7

x + x^{2} = 12

x^{2} - 16 x - 84 = - 5

7 x^{2} + 21 x - 28 = 0

2 = 2 x^{2}

What are the solutions to the equation 0=x^2+2x-48 ?
The solutions to the equation 0=x^2+2x-48 are x=6,x=-8
Find the zeros of 0=x^2+2x-48
The zeros of 0=x^2+2x-48 are x=6,x=-8