Question

valor de n(n+10).12=300???????????????????​

Answer 1

Resposta:

n(n+10).12=300

n(n+10)=300/12

n(n+10)=25

n²+10n-25=0

$\displaystyle\\Aplicando~a~f\'{o}rmula~de~Bhaskara~para~n^{2}+10n-25=0~~\\e~comparando~com~(a)n^{2}+(b)n+(c)=0,~temos~a=1{;}~b=10~e~c=-25\\\Delta=(b)^{2}-4(a)(c)=(10)^{2}-4(1)(-25)=100-(-100)=200\\\sqrt{\Delta}=\sqrt{200} =\sqrt{2.100} =\sqrt{2}.\sqrt{100} =10\sqrt{2} \\\\n^{'}=\frac{-(10)-\sqrt{200}}{2(1)}=\frac{-10-10\sqrt{2}}{2}=-5-5\sqrt{2}\\\\n^{''}=\frac{-(10)+\sqrt{200}}{2(1)}=\frac{-10+10\sqrt{2}}{2}=-5+5\sqrt{2} \\\\\\S=\{-5-5\sqrt{2},~-5+5\sqrt{2}\}$