Question

При каких значениях x верно равенство: (4x+5)^2=5x^2+4x?

Answer 1

$(4x+5)^2=5x^2+4x$
$16x^2+25+40x-5x^2-4x=0$
$11x^2+36x+25=0$
$D=b^2-4ac=1296-1100=196=14^2$
$x_{1} = \frac{-b+\sqrt{D}}{2a} =\frac{-36+14}{22}=-1$
$x_{2} = \frac{-b-\sqrt{D}}{2a} =\frac{-36-14}{22}= \frac{-50}{22} =-2 \frac{3}{11}$