What are the solutions to the equation -2x^2+120x+500=0 ?

The solutions to the equation -2x^2+120x+500=0 are x=-5(sqrt(46)-6),x=5(6+sqrt(46))

Find the zeros of -2x^2+120x+500=0

The zeros of -2x^2+120x+500=0 are x=-5(sqrt(46)-6),x=5(6+sqrt(46))

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- 2 x^{2} + 120 x + 500 = 0

x = - 5 (46 - 6), x = 5 (6 + 46)

Decimal Notation

x = - 3.91164 \dots, x = 63.91164 \dots

Solution steps

- 2 x^{2} + 120 x + 500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 2 ) \cdot 500}{2 ( - 2 )}

12 0^{2} - 4 (- 2) \cdot 500 = 2046

x_{1, 2} = \frac{- 120 \pm 20 46}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 120 + 20 46}{2 ( - 2 )}, x_{2} = \frac{- 120 - 20 46}{2 ( - 2 )}

x = \frac{- 120 + 20 46}{2 ( - 2 )} : - 5 (46 - 6)

x = \frac{- 120 - 20 46}{2 ( - 2 )} : 5 (6 + 46)

The solutions to the quadratic equation are:

x = - 5 (46 - 6), x = 5 (6 + 46)

x = 3 x - x^{2}

6 x^{2} - 48 = 0

x, 3 x^{2} + 5 ln (y) = 12

m^{2} + 4 m - 4 = 0

3 (x - 2)^{2} + 8 = 2

What are the solutions to the equation -2x^2+120x+500=0 ?
The solutions to the equation -2x^2+120x+500=0 are x=-5(sqrt(46)-6),x=5(6+sqrt(46))
Find the zeros of -2x^2+120x+500=0
The zeros of -2x^2+120x+500=0 are x=-5(sqrt(46)-6),x=5(6+sqrt(46))