What are the solutions to the equation x^2+4x=98 ?

The solutions to the equation x^2+4x=98 are x=-2+sqrt(102),x=-2-sqrt(102)

Find the zeros of x^2+4x=98

The zeros of x^2+4x=98 are x=-2+sqrt(102),x=-2-sqrt(102)

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x^{2} + 4 x = 98

x = - 2 + 102, x = - 2 - 102

Solution steps

x^{2} + 4 x = 98

Move

98

to the left side

x^{2} + 4 x - 98 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 98 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 98) = 2102

x_{1, 2} = \frac{- 4 \pm 2 102}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 102}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 102}{2 \cdot 1}

x = \frac{- 4 + 2 102}{2 \cdot 1} : - 2 + 102

x = \frac{- 4 - 2 102}{2 \cdot 1} : - 2 - 102

The solutions to the quadratic equation are:

x = - 2 + 102, x = - 2 - 102

x^{2} + 21 = 142

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

3 x^{2} = 2 - 5 x

4 x^{2} - 10 x - 4 = 0

n^{2} - 10 = 26

What are the solutions to the equation x^2+4x=98 ?
The solutions to the equation x^2+4x=98 are x=-2+sqrt(102),x=-2-sqrt(102)
Find the zeros of x^2+4x=98
The zeros of x^2+4x=98 are x=-2+sqrt(102),x=-2-sqrt(102)