Question

х^2+4х+(m-12)=0 помогите решить уровнение

Answer 1

D=4^2-4*(m-12)=16-4m-48=-4m-32=-4(m+8)
x1=(-4+(-4(m+8))/2=-4(-1+m+8)=-4(m+7)
x2=(-4-(-4(m+8))/2=-4(1+m+8)=-4(m+9)
ответ:-4(m+7);-4(m+9)

Answer 2

-4*(m-12)= - 4m+48

Answer 3

$x^{2} +4x+(m-12)=0 \ \ D=b^{2} -4ac \ x_{1}= frac{-b- sqrt{D} }{2a} \ x_{2}= frac{-b+ sqrt{D} }{2a} \ \ D=4^{2}-4(m-12)=16-4m+48=64-4m=4(16-m) \ \ x_{1}= frac{-4- sqrt{4(16-m)} }{2*1} = frac{-4- 2sqrt{16-m} }{2} = frac{-6sqrt{16-m} }{2} = -3sqrt{16-m}\ x_{2}= frac{-4+sqrt{4(16-m)} }{2*1} = frac{-4 + 2sqrt{16-m} }{2} = frac{-2sqrt{16-m} }{2} = -sqrt{16-m}$