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

The solutions to the equation 5x^2-x-7=0 are x=(1+sqrt(141))/(10),x=(1-sqrt(141))/(10)

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

The zeros of 5x^2-x-7=0 are x=(1+sqrt(141))/(10),x=(1-sqrt(141))/(10)

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

x = \frac{1 + 141}{10}, x = \frac{1 - 141}{10}

Decimal Notation

x = 1.28743 \dots, x = - 1.08743 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 5 (- 7) = 141

x_{1, 2} = \frac{- ( - 1 ) \pm 141}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 141}{2 \cdot 5}, x_{2} = \frac{- ( - 1 ) - 141}{2 \cdot 5}

x = \frac{- ( - 1 ) + 141}{2 \cdot 5} : \frac{1 + 141}{10}

x = \frac{- ( - 1 ) - 141}{2 \cdot 5} : \frac{1 - 141}{10}

The solutions to the quadratic equation are:

x = \frac{1 + 141}{10}, x = \frac{1 - 141}{10}

\frac{x}{- 5} = 6

∣ 4 - x ∣ + ∣ 2 x - 5 ∣ \geq 7 - x

h^{2} + 48 = 16 h

2 a^{2} + 7 ab - 15 b^{2}

200 \cdot \frac{π}{12} \cdot \frac{1.16}{1.3} \cdot 5

What are the solutions to the equation 5x^2-x-7=0 ?
The solutions to the equation 5x^2-x-7=0 are x=(1+sqrt(141))/(10),x=(1-sqrt(141))/(10)
Find the zeros of 5x^2-x-7=0
The zeros of 5x^2-x-7=0 are x=(1+sqrt(141))/(10),x=(1-sqrt(141))/(10)