What are the solutions to the equation 7g^2-7g-5=0 ?

The solutions to the equation 7g^2-7g-5=0 are g=(7+3sqrt(21))/(14),g=(7-3sqrt(21))/(14)

Find the zeros of 7g^2-7g-5=0

The zeros of 7g^2-7g-5=0 are g=(7+3sqrt(21))/(14),g=(7-3sqrt(21))/(14)

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7 g^{2} - 7 g - 5 = 0

g = \frac{7 + 3 21}{14}, g = \frac{7 - 3 21}{14}

Solution steps

7 g^{2} - 7 g - 5 = 0

Solve with the quadratic formula

g_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 7 ( - 5 )}{2 \cdot 7}

(- 7)^{2} - 4 \cdot 7 (- 5) = 321

g_{1, 2} = \frac{- ( - 7 ) \pm 3 21}{2 \cdot 7}

Separate the solutions

g_{1} = \frac{- ( - 7 ) + 3 21}{2 \cdot 7}, g_{2} = \frac{- ( - 7 ) - 3 21}{2 \cdot 7}

g = \frac{- ( - 7 ) + 3 21}{2 \cdot 7} : \frac{7 + 3 21}{14}

g = \frac{- ( - 7 ) - 3 21}{2 \cdot 7} : \frac{7 - 3 21}{14}

The solutions to the quadratic equation are:

g = \frac{7 + 3 21}{14}, g = \frac{7 - 3 21}{14}

x^{2} - 3 x - 2.5 = 0

3 p (g + 5) - 2 (g + 5)

r - 11 > - 17

(\frac{16}{3})^{3} - 8 (\frac{16}{3})^{2}

(x + y)^{3} (3 x - 2 y)^{4} + 2 (x + y)^{4} (3 x - 2 y)^{3}

What are the solutions to the equation 7g^2-7g-5=0 ?
The solutions to the equation 7g^2-7g-5=0 are g=(7+3sqrt(21))/(14),g=(7-3sqrt(21))/(14)
Find the zeros of 7g^2-7g-5=0
The zeros of 7g^2-7g-5=0 are g=(7+3sqrt(21))/(14),g=(7-3sqrt(21))/(14)