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

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

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

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

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

q = - 344, q = 12

Solution steps

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

Multiply both sides by

100

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

Move

46800

to the left side

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

Move

8300 q

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

q = - 344, q = 12

x^{2} + 2 x - 3 = x + 3

x^{2} = 3333

(z + 3)^{2} + (z - 3)^{2} = 10

x^{2} + x + 29 = - 9 x + 8

4 x - 7 = x^{2} - 2 x + 2

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