Question

(2x+3)^2-(x-5)^2=0
помогите решить​

Answer 1

Ответ:

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Answer 2

$\displaystyle(2x+3)^{2} -(x-5)^{2} =0\\4x^{2} +12x+9-(x^{2} -10x+25)=0\\4x^{2} +12x+9-x^{2} +10x-25=0\\3x^{2} +22x-16=0\\D=b^{2} -4ac=484-4\times3\times(-16)=484+192=676\\x_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{-22+26}{2\times3} =\frac{4}{6} =\frac{2}{3} \\x_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{-22-26}{2\times3} =-\frac{48}{6} =-8$