What are the solutions to the equation 7x^2+41x-6=0 ?

The solutions to the equation 7x^2+41x-6=0 are x= 1/7 ,x=-6

Find the zeros of 7x^2+41x-6=0

The zeros of 7x^2+41x-6=0 are x= 1/7 ,x=-6

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7 x^{2} + 41 x - 6 = 0

x = \frac{1}{7}, x = - 6

Decimal Notation

x = 0.14285 \dots, x = - 6

Solution steps

7 x^{2} + 41 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 41 \pm 4 1 ^{2} - 4 \cdot 7 ( - 6 )}{2 \cdot 7}

4 1^{2} - 4 \cdot 7 (- 6) = 43

x_{1, 2} = \frac{- 41 \pm 43}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 41 + 43}{2 \cdot 7}, x_{2} = \frac{- 41 - 43}{2 \cdot 7}

x = \frac{- 41 + 43}{2 \cdot 7} : \frac{1}{7}

x = \frac{- 41 - 43}{2 \cdot 7} : - 6

The solutions to the quadratic equation are:

x = \frac{1}{7}, x = - 6

z, x^{2} + y^{2} + (z - 2)^{2} = 4

x, 4 y^{2} + 4 z^{2} - x^{2} - 16 = 0

- \frac{15}{4} t^{2} - 15 t + 3000 = 0

6 x^{2} - 2 x - 3 = 0

8 + 0.4 q = 200 - 0.2 q^{2}

What are the solutions to the equation 7x^2+41x-6=0 ?
The solutions to the equation 7x^2+41x-6=0 are x= 1/7 ,x=-6
Find the zeros of 7x^2+41x-6=0
The zeros of 7x^2+41x-6=0 are x= 1/7 ,x=-6