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Last updated at May 29, 2018 by Teachoo
Misc 9 Solve the equation 21π₯2 β 28π₯ + 10 = 0 21π₯2 β 28π₯ + 10 = 0 The Above equation of the form ππ₯2 + ππ₯ + π = 0 Where a = 21, b = β28, and c = 10 π₯ = (βπ Β± β( π^2 β 4ππ ))/2π π₯ = (β ( β 28) Β± β((β28)^2 β4 Γ 21 Γ10))/(2 Γ 21) = (28 Β± β(784 β 840))/42 = (28 Β± β(β56))/42 = (28 Β± β(β 14 Γ4 ))/42 = (28 Β± 2β(β 14))/42 = (2 [14 Β± β(β 14) ])/42 = (2 [ 14 Β± β(β 14) ])/(2 Γ 21) = ( [ 14 Β± β(β 14) ] )/21 = ( 14 Β± β(β 1 Γ 14) )/21 = ( 14 Β± β(β 1) Γβ14)/21 = ( 14 Β± β14 π)/21 Thus, π₯ = ( 14 Β± β14 π)/21