What are the solutions to the equation -0.25x^2+1500=83x+468 ?

The solutions to the equation -0.25x^2+1500=83x+468 are x=-344,x=12

Find the zeros of -0.25x^2+1500=83x+468

The zeros of -0.25x^2+1500=83x+468 are x=-344,x=12

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- 0.25 x^{2} + 1500 = 83 x + 468

x = - 344, x = 12

Solution steps

- 0.25 x^{2} + 1500 = 83 x + 468

Multiply both sides by

100

- 25 x^{2} + 150000 = 8300 x + 46800

Move

46800

to the left side

- 25 x^{2} + 103200 = 8300 x

Move

8300 x

to the left side

- 25 x^{2} + 103200 - 8300 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 25 x^{2} - 8300 x + 103200 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8300 ) \pm ( - 8300 ) ^{2} - 4 ( - 25 ) \cdot 103200}{2 ( - 25 )}

(- 8300)^{2} - 4 (- 25) \cdot 103200 = 8900

x_{1, 2} = \frac{- ( - 8300 ) \pm 8900}{2 ( - 25 )}

Separate the solutions

x_{1} = \frac{- ( - 8300 ) + 8900}{2 ( - 25 )}, x_{2} = \frac{- ( - 8300 ) - 8900}{2 ( - 25 )}

x = \frac{- ( - 8300 ) + 8900}{2 ( - 25 )} : - 344

x = \frac{- ( - 8300 ) - 8900}{2 ( - 25 )} : 12

The solutions to the quadratic equation are:

x = - 344, x = 12

2 y = y^{2} - 3

x, x^{2} + 4 x - 32 = 0

- 2 x^{2} + 4 x + 48 = 0

(3 x - 4)^{2} + (x - 3) (x - 1) = 10

x, f = m^{2} x^{2} + 2 m x + 1

What are the solutions to the equation -0.25x^2+1500=83x+468 ?
The solutions to the equation -0.25x^2+1500=83x+468 are x=-344,x=12
Find the zeros of -0.25x^2+1500=83x+468
The zeros of -0.25x^2+1500=83x+468 are x=-344,x=12