What are the solutions to the equation 3600+25x+1/2 x^2=(175-1/2 x)x ?

The solutions to the equation 3600+25x+1/2 x^2=(175-1/2 x)x are x=120,x=30

Find the zeros of 3600+25x+1/2 x^2=(175-1/2 x)x

The zeros of 3600+25x+1/2 x^2=(175-1/2 x)x are x=120,x=30

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3600 + 25 x + \frac{1}{2} x^{2} = (175 - \frac{1}{2} x) x

x = 120, x = 30

Solution steps

3600 + 25 x + \frac{1}{2} x^{2} = (175 - \frac{1}{2} x) x

Multiply both sides by

2

7200 + 50 x + x^{2} = 2 x (175 - \frac{1}{2} x)

Expand

2 x (175 - \frac{1}{2} x) : 350 x - x^{2}

7200 + 50 x + x^{2} = 350 x - x^{2}

Move

x^{2}

to the left side

7200 + 50 x + 2 x^{2} = 350 x

Move

350 x

to the left side

2 x^{2} - 300 x + 7200 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 300 ) \pm ( - 300 ) ^{2} - 4 \cdot 2 \cdot 7200}{2 \cdot 2}

(- 300)^{2} - 4 \cdot 2 \cdot 7200 = 180

x_{1, 2} = \frac{- ( - 300 ) \pm 180}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 300 ) + 180}{2 \cdot 2}, x_{2} = \frac{- ( - 300 ) - 180}{2 \cdot 2}

x = \frac{- ( - 300 ) + 180}{2 \cdot 2} : 120

x = \frac{- ( - 300 ) - 180}{2 \cdot 2} : 30

The solutions to the quadratic equation are:

x = 120, x = 30

4 x^{2} = 2 - 7 x

x^{2} + (x + 5)^{2} + 83 = 1200

x^{2} + 35 = - 12 x

t^{2} + 6 t + 5 = 0

z^{2} - 6 z

What are the solutions to the equation 3600+25x+1/2 x^2=(175-1/2 x)x ?
The solutions to the equation 3600+25x+1/2 x^2=(175-1/2 x)x are x=120,x=30
Find the zeros of 3600+25x+1/2 x^2=(175-1/2 x)x
The zeros of 3600+25x+1/2 x^2=(175-1/2 x)x are x=120,x=30