What are the solutions to the equation 200x-2x^2=-x^2+10x+4125 ?

The solutions to the equation 200x-2x^2=-x^2+10x+4125 are x=165,x=25

Find the zeros of 200x-2x^2=-x^2+10x+4125

The zeros of 200x-2x^2=-x^2+10x+4125 are x=165,x=25

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200 x - 2 x^{2} = - x^{2} + 10 x + 4125

x = 165, x = 25

Solution steps

200 x - 2 x^{2} = - x^{2} + 10 x + 4125

Switch sides

- x^{2} + 10 x + 4125 = 200 x - 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} + 10 x + 4125 = 200 x

Move

200 x

to the left side

x^{2} - 190 x + 4125 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 190 ) \pm ( - 190 ) ^{2} - 4 \cdot 1 \cdot 4125}{2 \cdot 1}

(- 190)^{2} - 4 \cdot 1 \cdot 4125 = 140

x_{1, 2} = \frac{- ( - 190 ) \pm 140}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 190 ) + 140}{2 \cdot 1}, x_{2} = \frac{- ( - 190 ) - 140}{2 \cdot 1}

x = \frac{- ( - 190 ) + 140}{2 \cdot 1} : 165

x = \frac{- ( - 190 ) - 140}{2 \cdot 1} : 25

The solutions to the quadratic equation are:

x = 165, x = 25

- x (2 x + 1) \geq - 3

5 x^{2} + 5 x + 8 = 0

∣ x + 2 ∣ < 3 ∣ x ∣

16 = \frac{5}{4} (6 x + 20)

0. 2^{3}

What are the solutions to the equation 200x-2x^2=-x^2+10x+4125 ?
The solutions to the equation 200x-2x^2=-x^2+10x+4125 are x=165,x=25
Find the zeros of 200x-2x^2=-x^2+10x+4125
The zeros of 200x-2x^2=-x^2+10x+4125 are x=165,x=25