What are the solutions to the equation 0.5x^2+2x=120 ?

The solutions to the equation 0.5x^2+2x=120 are x=2(sqrt(61)-1),x=-2(1+sqrt(61))

Find the zeros of 0.5x^2+2x=120

The zeros of 0.5x^2+2x=120 are x=2(sqrt(61)-1),x=-2(1+sqrt(61))

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0.5 x^{2} + 2 x = 120

x = 2 (61 - 1), x = - 2 (1 + 61)

Solution steps

0.5 x^{2} + 2 x = 120

Multiply both sides by

10

5 x^{2} + 20 x = 1200

Move

1200

to the left side

5 x^{2} + 20 x - 1200 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 5 ( - 1200 )}{2 \cdot 5}

2 0^{2} - 4 \cdot 5 (- 1200) = 2061

x_{1, 2} = \frac{- 20 \pm 20 61}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 20 + 20 61}{2 \cdot 5}, x_{2} = \frac{- 20 - 20 61}{2 \cdot 5}

x = \frac{- 20 + 20 61}{2 \cdot 5} : 2 (61 - 1)

x = \frac{- 20 - 20 61}{2 \cdot 5} : - 2 (1 + 61)

The solutions to the quadratic equation are:

x = 2 (61 - 1), x = - 2 (1 + 61)

2 x^{2} - 14 x + 9 = (x - 4)^{2}

g (- 2) = \frac{- 2 - 4}{- 2 + 10 - 36}

- 2 (x - 5) + 4 x = x - 4

x^{2} + 39 x + 360

\frac{9 - x}{11} = - 4

What are the solutions to the equation 0.5x^2+2x=120 ?
The solutions to the equation 0.5x^2+2x=120 are x=2(sqrt(61)-1),x=-2(1+sqrt(61))
Find the zeros of 0.5x^2+2x=120
The zeros of 0.5x^2+2x=120 are x=2(sqrt(61)-1),x=-2(1+sqrt(61))