What are the solutions to the equation 2x^2+8x-7=0 ?

The solutions to the equation 2x^2+8x-7=0 are x=(-4+sqrt(30))/2 ,x=-(4+sqrt(30))/2

Find the zeros of 2x^2+8x-7=0

The zeros of 2x^2+8x-7=0 are x=(-4+sqrt(30))/2 ,x=-(4+sqrt(30))/2

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2 x^{2} + 8 x - 7 = 0

x = \frac{- 4 + 30}{2}, x = - \frac{4 + 30}{2}

Decimal Notation

x = 0.73861 \dots, x = - 4.73861 \dots

Solution steps

2 x^{2} + 8 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 7) = 230

x_{1, 2} = \frac{- 8 \pm 2 30}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 2 30}{2 \cdot 2}, x_{2} = \frac{- 8 - 2 30}{2 \cdot 2}

x = \frac{- 8 + 2 30}{2 \cdot 2} : \frac{- 4 + 30}{2}

x = \frac{- 8 - 2 30}{2 \cdot 2} : - \frac{4 + 30}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 30}{2}, x = - \frac{4 + 30}{2}

4 x = x^{2} - 12

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

x = \frac{9 ( - 3 ) - 2 ( - 2 )}{15 ( - 3 ) - 3 ( - 2 )}

8 - \frac{x}{3} \leq 7

(- 6)^{- 2}

What are the solutions to the equation 2x^2+8x-7=0 ?
The solutions to the equation 2x^2+8x-7=0 are x=(-4+sqrt(30))/2 ,x=-(4+sqrt(30))/2
Find the zeros of 2x^2+8x-7=0
The zeros of 2x^2+8x-7=0 are x=(-4+sqrt(30))/2 ,x=-(4+sqrt(30))/2