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

The solutions to the equation 5x^2-5x-7=0 are x=(5+sqrt(165))/(10),x=(5-sqrt(165))/(10)

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

The zeros of 5x^2-5x-7=0 are x=(5+sqrt(165))/(10),x=(5-sqrt(165))/(10)

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

x = \frac{5 + 165}{10}, x = \frac{5 - 165}{10}

Decimal

x = 1.78452 \dots, x = - 0.78452 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 5 (- 7) = 165

x_{1, 2} = \frac{- ( - 5 ) \pm 165}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 165}{2 \cdot 5}, x_{2} = \frac{- ( - 5 ) - 165}{2 \cdot 5}

x = \frac{- ( - 5 ) + 165}{2 \cdot 5} : \frac{5 + 165}{10}

x = \frac{- ( - 5 ) - 165}{2 \cdot 5} : \frac{5 - 165}{10}

The solutions to the quadratic equation are:

x = \frac{5 + 165}{10}, x = \frac{5 - 165}{10}

4 d^{2} - 15 d + 16 = 0

- 4 + ∣ 7 - 2 k ∣ < - 19

∣ x + 4 ∣ = ∣ 2 x - 7 ∣

4 x^{4} + 4 x^{3} + 5 x^{2} + 4 x + 1

t^{2} + 6 t + 8

What are the solutions to the equation 5x^2-5x-7=0 ?
The solutions to the equation 5x^2-5x-7=0 are x=(5+sqrt(165))/(10),x=(5-sqrt(165))/(10)
Find the zeros of 5x^2-5x-7=0
The zeros of 5x^2-5x-7=0 are x=(5+sqrt(165))/(10),x=(5-sqrt(165))/(10)