What are the solutions to the equation 496.4=200+4x+0.1x^2 ?

The solutions to the equation 496.4=200+4x+0.1x^2 are x=38,x=-78

Find the zeros of 496.4=200+4x+0.1x^2

The zeros of 496.4=200+4x+0.1x^2 are x=38,x=-78

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496.4 = 200 + 4 x + 0.1 x^{2}

x = 38, x = - 78

Solution steps

496.4 = 200 + 4 x + 0.1 x^{2}

Multiply both sides by

10

4964 = 2000 + 40 x + x^{2}

Switch sides

2000 + 40 x + x^{2} = 4964

Move

4964

to the left side

x^{2} + 40 x - 2964 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 1 \cdot ( - 2964 )}{2 \cdot 1}

4 0^{2} - 4 \cdot 1 \cdot (- 2964) = 116

x_{1, 2} = \frac{- 40 \pm 116}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 40 + 116}{2 \cdot 1}, x_{2} = \frac{- 40 - 116}{2 \cdot 1}

x = \frac{- 40 + 116}{2 \cdot 1} : 38

x = \frac{- 40 - 116}{2 \cdot 1} : - 78

The solutions to the quadratic equation are:

x = 38, x = - 78

d + 6 < 2

n = 0.42 (2.47)^{4}

x^{4} \leq 9

- 3 y = 24

What are the solutions to the equation 496.4=200+4x+0.1x^2 ?
The solutions to the equation 496.4=200+4x+0.1x^2 are x=38,x=-78
Find the zeros of 496.4=200+4x+0.1x^2
The zeros of 496.4=200+4x+0.1x^2 are x=38,x=-78