What are the solutions to the equation 2x^2+6x-31=0 ?

The solutions to the equation 2x^2+6x-31=0 are x=(-3+sqrt(71))/2 ,x=-(3+sqrt(71))/2

Find the zeros of 2x^2+6x-31=0

The zeros of 2x^2+6x-31=0 are x=(-3+sqrt(71))/2 ,x=-(3+sqrt(71))/2

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2 x^{2} + 6 x - 31 = 0

x = \frac{- 3 + 71}{2}, x = - \frac{3 + 71}{2}

Decimal Notation

x = 2.71307 \dots, x = - 5.71307 \dots

Solution steps

2 x^{2} + 6 x - 31 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 31 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 31) = 271

x_{1, 2} = \frac{- 6 \pm 2 71}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 2 71}{2 \cdot 2}, x_{2} = \frac{- 6 - 2 71}{2 \cdot 2}

x = \frac{- 6 + 2 71}{2 \cdot 2} : \frac{- 3 + 71}{2}

x = \frac{- 6 - 2 71}{2 \cdot 2} : - \frac{3 + 71}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 71}{2}, x = - \frac{3 + 71}{2}

32 = 704 + 16 t - 16 t^{2}

\frac{1}{y} < 1

(- 18 m^{2} n)^{2} (- \frac{1}{6} m n^{2})

(\frac{64}{25})^{- \frac{1}{2}}

\frac{9 ^{12}}{9 ^{8}}

What are the solutions to the equation 2x^2+6x-31=0 ?
The solutions to the equation 2x^2+6x-31=0 are x=(-3+sqrt(71))/2 ,x=-(3+sqrt(71))/2
Find the zeros of 2x^2+6x-31=0
The zeros of 2x^2+6x-31=0 are x=(-3+sqrt(71))/2 ,x=-(3+sqrt(71))/2