Question

помогите решить уравнение
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Answer 1

X^2+4x=21
X^2+4x-21=0
D=16+84=100
x1=(-4+10)/2=3
x2=(-4-10)/2=-7

Answer 2

$\displaystyle \tt x^2+4x=21\\\displaystyle \tt x^2+4x-21=0\\\displaystyle \tt D=4^2-4\cdot1\cdot(-21)=16+84=100\\\displaystyle \tt \sqrt{D}=\sqrt{100}=10\\\displaystyle \tt x_1=\frac{-4+10}{2}=\frac{6}{2}=3\\\displaystyle \tt x_2=\frac{-4-10}{2}=\frac{-14}{2}=-7$