Question

Qual raíz da equação x²-14x+48=0

Answer 1

usei a fórmula direta de baskara
$x = \frac{- b + ou - \sqrt{ {b}^{2} - 4ac} }{2}$

Answer 2

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-14x+48=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-14~e~c=48\\\\\Delta=(b)^{2}-4(a)(c)=(-14)^{2}-4(1)(48)=196-(192)=4\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-14)-\sqrt{4}}{2(1)}=\frac{14-2}{2}=\frac{12}{2}=6\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-14)+\sqrt{4}}{2(1)}=\frac{14+2}{2}=\frac{16}{2}=8\\\\S=\{6,~8\}$