What are the solutions to the equation 428=-0.32q^2+500 ?

The solutions to the equation 428=-0.32q^2+500 are q=15,q=-15

Find the zeros of 428=-0.32q^2+500

The zeros of 428=-0.32q^2+500 are q=15,q=-15

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428 = - 0.32 q^{2} + 500

q = 15, q = - 15

Solution steps

428 = - 0.32 q^{2} + 500

Switch sides

- 0.32 q^{2} + 500 = 428

Move

500

to the right side

- 0.32 q^{2} = - 72

Divide both sides by

- 0.32

q^{2} = 225

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

q = 225, q = - 225

225 = 15

- 225 = - 15

q = 15, q = - 15

x^{2} + 7 x + 7 + 8 = 0

x^{2} + 14 x + 35 = - 5

(x) = 0.001 x^{2} - 5.98 x + 80

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

(x^{2} + 3) 2 = 4 x^{2} + 12

What are the solutions to the equation 428=-0.32q^2+500 ?
The solutions to the equation 428=-0.32q^2+500 are q=15,q=-15
Find the zeros of 428=-0.32q^2+500
The zeros of 428=-0.32q^2+500 are q=15,q=-15