Question

alguém pra me ajudar a resolver essa equação de 2°grau
$x { }^{2} - 8x - 48 = 0$
gostaria muito da ajuda de alguém
Obg desde já

Answer 1

a)
x² -8x -48=0

a=1
b=-8
c=-48

Δ=b²-4ac
Δ=(-8)²-4(1)(-48)
Δ=64+192
Δ=256

$x= \frac{-b\pm \sqrt{\Delta} }{2a} =~~ \frac{-(-8)\pm \sqrt{256} }{2} = \frac{8\pm16}{2} \\ \\ x'= \frac{8+16}{2} = \frac{24}{2} =12 \\ \\ x"= \frac{8-16}{2}=- \frac{8}{2} =-4 \\ \\ S=\{-4,12\}$ 

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$b) \\ x^{2} -14x+45=0 \\ a=1 \\ b=-14 \\ c=45 \\ \\ \Delta=b^2-4ac \\ \Delta=(-14)^2-4(1)(45) \\ \Delta=196-180 \\ \Delta=16 \\ \\ x= \frac{-b\pm \sqrt{\Delta} }{2a} = \frac{-(-14)\pm \sqrt{16} }{2} = \frac{14\pm4}{2} \\ \\ x'= \frac{14+4}{2} = \frac{18}{2} =9 \\ \\ x"= \frac{14-4}{2} = \frac{10}{2} =5 \\ \\ S=\{5,9\}$ 

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$c)~~ x^{2} +6x+9=0 \\ \\ a=1 \\ b=6 \\ c=9 \\ \Delta=b^2-4ac \\ \Delta=6^2-4(1)(9) \\ \Delta=36-36 \\ \Delta=0 \\ \\ x= \frac{-b\pm \sqrt{\Delta} }{2a} = \frac{-6\pm \sqrt{0} }{2} = \frac{-6\pm0}{2} \\ \\ x'=x"=- \frac{6}{2} =-3 \\ \\ S=\{-3\}$