What are the solutions to the equation 11x^2-26x-61=0 ?

The solutions to the equation 11x^2-26x-61=0 are x=(13+2sqrt(210))/(11),x=(13-2sqrt(210))/(11)

Find the zeros of 11x^2-26x-61=0

The zeros of 11x^2-26x-61=0 are x=(13+2sqrt(210))/(11),x=(13-2sqrt(210))/(11)

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11 x^{2} - 26 x - 61 = 0

x = \frac{13 + 2 210}{11}, x = \frac{13 - 2 210}{11}

Solution steps

11 x^{2} - 26 x - 61 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 26 ) \pm ( - 26 ) ^{2} - 4 \cdot 11 ( - 61 )}{2 \cdot 11}

(- 26)^{2} - 4 \cdot 11 (- 61) = 4210

x_{1, 2} = \frac{- ( - 26 ) \pm 4 210}{2 \cdot 11}

Separate the solutions

x_{1} = \frac{- ( - 26 ) + 4 210}{2 \cdot 11}, x_{2} = \frac{- ( - 26 ) - 4 210}{2 \cdot 11}

x = \frac{- ( - 26 ) + 4 210}{2 \cdot 11} : \frac{13 + 2 210}{11}

x = \frac{- ( - 26 ) - 4 210}{2 \cdot 11} : \frac{13 - 2 210}{11}

The solutions to the quadratic equation are:

x = \frac{13 + 2 210}{11}, x = \frac{13 - 2 210}{11}

- 6 < - \frac{2}{5} (1 - x) \leq 4

2 (4) - 3 (x - 1) = - 1

(\frac{3 a}{b})^{4}

\frac{3 r ^{2} p ^{6}}{t ^{4}} + \frac{4 r ^{2} p ^{6}}{t ^{4}}

(x - 2)^{2} (x + 1)^{3} (x + 5) \leq 0

What are the solutions to the equation 11x^2-26x-61=0 ?
The solutions to the equation 11x^2-26x-61=0 are x=(13+2sqrt(210))/(11),x=(13-2sqrt(210))/(11)
Find the zeros of 11x^2-26x-61=0
The zeros of 11x^2-26x-61=0 are x=(13+2sqrt(210))/(11),x=(13-2sqrt(210))/(11)