a)
[tex]x^2+10x+24=0[/tex]
[tex]\frac{-10+\sqrt{10^2-4\cdot \:1\cdot \:24}}{2\cdot \:1} = \\[/tex]
[tex]\sqrt{10^2-4\cdot \:1\cdot \:24}=\sqrt{4}[/tex]
[tex]=\frac{-10+\sqrt{4}}{2\cdot \:1} =\frac{-10+\sqrt{4}}{2} =\frac{-10+2}{2} =\frac{-8}{2} =-\frac{8}{2} =-4[/tex]
[tex]\frac{-10-\sqrt{10^2-4\cdot \:1\cdot \:24}}{2\cdot \:1} =\frac{-10-\sqrt{4}}{2\cdot \:1} =\frac{-10-\sqrt{4}}{2} =\frac{-10-2}{2} =\frac{-12}{2} =-\frac{12}{2} =-6[/tex]
[tex]x=-4,\:x=-6[/tex]
b)
[tex]\left(x+2\right)^2=7-4\sqrt{3}[/tex]
[tex]x+2=\sqrt{7-4\sqrt{3}}[/tex]
[tex]\sqrt{7-4\sqrt{3}} =\sqrt{3-4\sqrt{3}+4} =\sqrt{1\cdot \:3-4\sqrt{3}+4}=\sqrt{\left(\sqrt{1}\right)^2\left(\sqrt{3}\right)^2-4\sqrt{3}+\left(\sqrt{4}\right)^2} =\sqrt{1^2\left(\sqrt{3}\right)^2-4\sqrt{3}+2^2} =\sqrt{1^2\left(\sqrt{3}\right)^2-2\cdot \:1\cdot \:2\sqrt{3}+2^2} =\sqrt{\left(1\cdot \sqrt{3}-2\right)^2} =\sqrt{\left(2-1\cdot \sqrt{3}\right)^2} =2-1\cdot \sqrt{3} =2-\sqrt{3}[/tex]
[tex]x+2=2-\sqrt{3}[/tex]
[tex]x+2-2=2-\sqrt{3}-2[/tex]
[tex]x=-\sqrt{3}[/tex]
[tex]x+2=-\sqrt{7-4\sqrt{3}}[/tex]
[tex]-\sqrt{7-4\sqrt{3}} =-\left(2-1\cdot \sqrt{3}\right) =-\left(2-\sqrt{3}\right)[/tex]
[tex]x+2-2=-\left(2-\sqrt{3}\right)-2[/tex]
[tex]x=\sqrt{3}-4[/tex]
[tex]x=-\sqrt{3},\:x=\sqrt{3}-4[/tex]
Sper că te-am ajutat!
Succes!
