What are the solutions to the equation 2x^2+4x-39=x^2 ?

The solutions to the equation 2x^2+4x-39=x^2 are x=-2+sqrt(43),x=-2-sqrt(43)

Find the zeros of 2x^2+4x-39=x^2

The zeros of 2x^2+4x-39=x^2 are x=-2+sqrt(43),x=-2-sqrt(43)

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2 x^{2} + 4 x - 39 = x^{2}

x = - 2 + 43, x = - 2 - 43

Decimal Notation

x = 4.55743 \dots, x = - 8.55743 \dots

Solution steps

2 x^{2} + 4 x - 39 = x^{2}

Move

x^{2}

to the left side

x^{2} + 4 x - 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 39 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 39) = 243

x_{1, 2} = \frac{- 4 \pm 2 43}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 43}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 43}{2 \cdot 1}

x = \frac{- 4 + 2 43}{2 \cdot 1} : - 2 + 43

x = \frac{- 4 - 2 43}{2 \cdot 1} : - 2 - 43

The solutions to the quadratic equation are:

x = - 2 + 43, x = - 2 - 43

x^{2} - 14 x + 4 = - 10 x + 9

x, x^{2} + x y + y^{2} = c

x^{2} - x + 1 = 2

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

1 + 0.2 x + 0.7 x^{2} = 0

What are the solutions to the equation 2x^2+4x-39=x^2 ?
The solutions to the equation 2x^2+4x-39=x^2 are x=-2+sqrt(43),x=-2-sqrt(43)
Find the zeros of 2x^2+4x-39=x^2
The zeros of 2x^2+4x-39=x^2 are x=-2+sqrt(43),x=-2-sqrt(43)