What are the solutions to the equation 80x+1/2 (4)x^2-1000=0 ?

The solutions to the equation 80x+1/2 (4)x^2-1000=0 are x=10,x=-50

Find the zeros of 80x+1/2 (4)x^2-1000=0

The zeros of 80x+1/2 (4)x^2-1000=0 are x=10,x=-50

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80 x + \frac{1}{2} (4) x^{2} - 1000 = 0

x = 10, x = - 50

Solution steps

80 x + \frac{1}{2} (4) x^{2} - 1000 = 0

Multiply both sides by

2

4 x^{2} + 160 x - 2000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 160 \pm 16 0 ^{2} - 4 \cdot 4 ( - 2000 )}{2 \cdot 4}

16 0^{2} - 4 \cdot 4 (- 2000) = 240

x_{1, 2} = \frac{- 160 \pm 240}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 160 + 240}{2 \cdot 4}, x_{2} = \frac{- 160 - 240}{2 \cdot 4}

x = \frac{- 160 + 240}{2 \cdot 4} : 10

x = \frac{- 160 - 240}{2 \cdot 4} : - 50

The solutions to the quadratic equation are:

x = 10, x = - 50

2 b^{2} = 10

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

y^{2} - 225 = 0

(z + 9)^{2} = - 9

2 x^{2} - 0.41 x - 1.27 = 0

What are the solutions to the equation 80x+1/2 (4)x^2-1000=0 ?
The solutions to the equation 80x+1/2 (4)x^2-1000=0 are x=10,x=-50
Find the zeros of 80x+1/2 (4)x^2-1000=0
The zeros of 80x+1/2 (4)x^2-1000=0 are x=10,x=-50