What are the solutions to the equation 13x^2-160x-284=0 ?

The solutions to the equation 13x^2-160x-284=0 are x=(2(40+29sqrt(3)))/(13),x=(2(40-29sqrt(3)))/(13)

Find the zeros of 13x^2-160x-284=0

The zeros of 13x^2-160x-284=0 are x=(2(40+29sqrt(3)))/(13),x=(2(40-29sqrt(3)))/(13)

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13 x^{2} - 160 x - 284 = 0

x = \frac{2 ( 40 + 29 3 )}{13}, x = \frac{2 ( 40 - 29 3 )}{13}

Decimal Notation

x = 13.88145 \dots, x = - 1.57376 \dots

Solution steps

13 x^{2} - 160 x - 284 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 160 ) \pm ( - 160 ) ^{2} - 4 \cdot 13 ( - 284 )}{2 \cdot 13}

(- 160)^{2} - 4 \cdot 13 (- 284) = 1163

x_{1, 2} = \frac{- ( - 160 ) \pm 116 3}{2 \cdot 13}

Separate the solutions

x_{1} = \frac{- ( - 160 ) + 116 3}{2 \cdot 13}, x_{2} = \frac{- ( - 160 ) - 116 3}{2 \cdot 13}

x = \frac{- ( - 160 ) + 116 3}{2 \cdot 13} : \frac{2 ( 40 + 29 3 )}{13}

x = \frac{- ( - 160 ) - 116 3}{2 \cdot 13} : \frac{2 ( 40 - 29 3 )}{13}

The solutions to the quadratic equation are:

x = \frac{2 ( 40 + 29 3 )}{13}, x = \frac{2 ( 40 - 29 3 )}{13}

\frac{5}{2} x + \frac{1}{16} = x + \frac{1}{4}

m^{2} + 13 m - 30

9 (z - 10) = - 4 z - 10

49 y^{2} + 14 y + 1

\frac{a}{12} = \frac{5}{6}

What are the solutions to the equation 13x^2-160x-284=0 ?
The solutions to the equation 13x^2-160x-284=0 are x=(2(40+29sqrt(3)))/(13),x=(2(40-29sqrt(3)))/(13)
Find the zeros of 13x^2-160x-284=0
The zeros of 13x^2-160x-284=0 are x=(2(40+29sqrt(3)))/(13),x=(2(40-29sqrt(3)))/(13)