What are the solutions to the equation 0=4x^2-260x+2000 ?

The solutions to the equation 0=4x^2-260x+2000 are x=(5(13+sqrt(89)))/2 ,x=(5(13-sqrt(89)))/2

Find the zeros of 0=4x^2-260x+2000

The zeros of 0=4x^2-260x+2000 are x=(5(13+sqrt(89)))/2 ,x=(5(13-sqrt(89)))/2

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0 = 4 x^{2} - 260 x + 2000

x = \frac{5 ( 13 + 89 )}{2}, x = \frac{5 ( 13 - 89 )}{2}

Decimal Notation

x = 56.08495 \dots, x = 8.91504 \dots

Solution steps

0 = 4 x^{2} - 260 x + 2000

Switch sides

4 x^{2} - 260 x + 2000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 260 ) \pm ( - 260 ) ^{2} - 4 \cdot 4 \cdot 2000}{2 \cdot 4}

(- 260)^{2} - 4 \cdot 4 \cdot 2000 = 2089

x_{1, 2} = \frac{- ( - 260 ) \pm 20 89}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 260 ) + 20 89}{2 \cdot 4}, x_{2} = \frac{- ( - 260 ) - 20 89}{2 \cdot 4}

x = \frac{- ( - 260 ) + 20 89}{2 \cdot 4} : \frac{5 ( 13 + 89 )}{2}

x = \frac{- ( - 260 ) - 20 89}{2 \cdot 4} : \frac{5 ( 13 - 89 )}{2}

The solutions to the quadratic equation are:

x = \frac{5 ( 13 + 89 )}{2}, x = \frac{5 ( 13 - 89 )}{2}

y, x^{3} + x^{2} y + 4 y^{2} = 6

x^{2} + 30 x + 7000 = 0

\frac{6}{7} x^{2} - x + \frac{2}{7} = 0

2 x^{2} + 4 = 3 x + 9

- 2 x^{2} + 4 x + 8 = 0

What are the solutions to the equation 0=4x^2-260x+2000 ?
The solutions to the equation 0=4x^2-260x+2000 are x=(5(13+sqrt(89)))/2 ,x=(5(13-sqrt(89)))/2
Find the zeros of 0=4x^2-260x+2000
The zeros of 0=4x^2-260x+2000 are x=(5(13+sqrt(89)))/2 ,x=(5(13-sqrt(89)))/2