Verified answer

[tex]\displaystyle\bf\\(5-x)^{2}=(4-x)(x+4)\\\\25-10x+x^{2} =16-x^{2} \\\\25-10x+x^{2} -16+x^{2} =0\\\\2x^{2} -10x+9=0\\\\D=(-10)^{2} -4\cdot 2\cdot 9=100-72=28=(2\sqrt{7} )^{2} \\\\\\x_{1} =\frac{10-2\sqrt{7} }{4} =\frac{5-\sqrt{7} }{2} =2,5-0,5\sqrt{7} \\\\\\x_{2} =\frac{10+2\sqrt{7} }{4} =\frac{5+\sqrt{7} }{2} =2,5+0,5\sqrt{7} \\\\\\Otvet \ : \ 2,5-0,5\sqrt{7} \ ; \ 2,5+0,5\sqrt{7}[/tex]