What are the solutions to the equation 2x^2+6x-7=2 ?

The solutions to the equation 2x^2+6x-7=2 are x=(3(sqrt(3)-1))/2 ,x=-(3(1+sqrt(3)))/2

Find the zeros of 2x^2+6x-7=2

The zeros of 2x^2+6x-7=2 are x=(3(sqrt(3)-1))/2 ,x=-(3(1+sqrt(3)))/2

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2 x^{2} + 6 x - 7 = 2

x = \frac{3 ( 3 - 1 )}{2}, x = - \frac{3 ( 1 + 3 )}{2}

Decimal Notation

x = 1.09807 \dots, x = - 4.09807 \dots

Solution steps

2 x^{2} + 6 x - 7 = 2

Move

2

to the left side

2 x^{2} + 6 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 9) = 63

x_{1, 2} = \frac{- 6 \pm 6 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 6 3}{2 \cdot 2}, x_{2} = \frac{- 6 - 6 3}{2 \cdot 2}

x = \frac{- 6 + 6 3}{2 \cdot 2} : \frac{3 ( 3 - 1 )}{2}

x = \frac{- 6 - 6 3}{2 \cdot 2} : - \frac{3 ( 1 + 3 )}{2}

The solutions to the quadratic equation are:

x = \frac{3 ( 3 - 1 )}{2}, x = - \frac{3 ( 1 + 3 )}{2}

x, 3 x^{2} + 30 x - 5 = 0

5 r^{2} + 24 r - 3 = 2

m, x^{2} + m (x)^{2} x (m (x) - 2) = 2 m (x)^{5}

9 (2 - x) = 2 x^{2}

2 c^{2} - 33 = 65

What are the solutions to the equation 2x^2+6x-7=2 ?
The solutions to the equation 2x^2+6x-7=2 are x=(3(sqrt(3)-1))/2 ,x=-(3(1+sqrt(3)))/2
Find the zeros of 2x^2+6x-7=2
The zeros of 2x^2+6x-7=2 are x=(3(sqrt(3)-1))/2 ,x=-(3(1+sqrt(3)))/2