jsteadg78p5y833
jsteadg78p5y833 jsteadg78p5y833
  • 02-11-2021
  • Mathematics
contestada

20x^2-32x=224 How do I solve this equation?

Respuesta :

anasweeney43 anasweeney43
  • 04-11-2021

Answer:

(x = 4/5 + 2/5 sqrt 74) or (x = 4/5 - 2/5 sqrt 74)

Step-by-step explanation:

Quadratic equation:

X = [ -b +- (sqrt b^2 - 4ac)] / 2a

20x^2 - 32 x - 224 = 0

a = 20, b = -32, and c = -224

X = - ( - 32 ) + - [sqrt -32^2 - 4(20)(-224)] /(2*20)

X = ( 32 + - sqrt 18944) / 40 (18944 = 2^8 times 74)

X = [32/40 + - (sqrt 2^8 * 74)] / 40

X = 4/5 + - (2^4 sqrt 74/40)

X = 4/5 + - 16/40 sqrt 74

(x = 4/5 + 2/5 sqrt 74) or (x = 4/5 - 2/5 sqrt 74)

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