What are the solutions to the equation 3x^2+26x+47=7 ?

The solutions to the equation 3x^2+26x+47=7 are x=-2,x=-20/3

Find the zeros of 3x^2+26x+47=7

The zeros of 3x^2+26x+47=7 are x=-2,x=-20/3

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3 x^{2} + 26 x + 47 = 7

x = - 2, x = - \frac{20}{3}

Decimal Notation

x = - 2, x = - 6.66666 \dots

Solution steps

3 x^{2} + 26 x + 47 = 7

Move

7

to the left side

3 x^{2} + 26 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 26 \pm 2 6 ^{2} - 4 \cdot 3 \cdot 40}{2 \cdot 3}

2 6^{2} - 4 \cdot 3 \cdot 40 = 14

x_{1, 2} = \frac{- 26 \pm 14}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 26 + 14}{2 \cdot 3}, x_{2} = \frac{- 26 - 14}{2 \cdot 3}

x = \frac{- 26 + 14}{2 \cdot 3} : - 2

x = \frac{- 26 - 14}{2 \cdot 3} : - \frac{20}{3}

The solutions to the quadratic equation are:

x = - 2, x = - \frac{20}{3}

2 x^{2} + 28 = 100

x^{2} - 20 x + 118 = 0

(x^{2} + 16) = 0

(6 x) (x - 6) = 0

- x^{2} - 10 x = 0

What are the solutions to the equation 3x^2+26x+47=7 ?
The solutions to the equation 3x^2+26x+47=7 are x=-2,x=-20/3
Find the zeros of 3x^2+26x+47=7
The zeros of 3x^2+26x+47=7 are x=-2,x=-20/3