What are the solutions to the equation 5300=100x^2-1200x+4000 ?

The solutions to the equation 5300=100x^2-1200x+4000 are x=13,x=-1

Find the zeros of 5300=100x^2-1200x+4000

The zeros of 5300=100x^2-1200x+4000 are x=13,x=-1

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5300 = 100 x^{2} - 1200 x + 4000

x = 13, x = - 1

Solution steps

5300 = 100 x^{2} - 1200 x + 4000

Switch sides

100 x^{2} - 1200 x + 4000 = 5300

Move

5300

to the left side

100 x^{2} - 1200 x - 1300 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1200 ) \pm ( - 1200 ) ^{2} - 4 \cdot 100 ( - 1300 )}{2 \cdot 100}

(- 1200)^{2} - 4 \cdot 100 (- 1300) = 1400

x_{1, 2} = \frac{- ( - 1200 ) \pm 1400}{2 \cdot 100}

Separate the solutions

x_{1} = \frac{- ( - 1200 ) + 1400}{2 \cdot 100}, x_{2} = \frac{- ( - 1200 ) - 1400}{2 \cdot 100}

x = \frac{- ( - 1200 ) + 1400}{2 \cdot 100} : 13

x = \frac{- ( - 1200 ) - 1400}{2 \cdot 100} : - 1

The solutions to the quadratic equation are:

x = 13, x = - 1

k^{2} + 2 k + 1 = 0

36 x^{2} - 4 x - \frac{35}{9}

x^{2} - 6 x + 30 = 0

1 - 2 x + x^{2} = 2 x - 2

x, 4 x^{2} + y^{2} - 8 x + 4 = 0

What are the solutions to the equation 5300=100x^2-1200x+4000 ?
The solutions to the equation 5300=100x^2-1200x+4000 are x=13,x=-1
Find the zeros of 5300=100x^2-1200x+4000
The zeros of 5300=100x^2-1200x+4000 are x=13,x=-1