Question

Помогите пожалуйста очень срочно нужно

Answer 1

Ответ:

Пошаговое объяснение:

3. $3.\\y^{2}  -12y +32 =0\\D= 144-128 = 16=(\sqrt{4} )^{2} \\ x_{1,2}= \frac{12(+-)4}{2}   \\\\x_{1}  = 4\\x_{2}  = 8\\\\\\4. \\x^{2} +5x-14=0\\\\x_{1,2}= \frac{-5(+-)9}{2}   \\\\\\x_{1}  = -7\\x_{2}  = 2\\\\\\\\5. \\x^{2} -6x-9=0\\\\x_{1,2}= \frac{6}{2}   \\\\\\x = 3\\\\6. \\25p^{2} -10p+1=0\\\\x_{1,2}= \frac{10}{2}   \\\\\\x = 5$

$7. \\3x^{2} +8x-3=x+6x^{2} \\\\ \\3x^{2} -7x+3=0 \\\\\\x_{1,2}= \frac{7(+-)\sqrt{13} }{6}   \\\\\\8. x^{2} +8x+16=3x+40\\x^{2} +5x-24=0\\\\\\x_{1,2}= \frac{-5(+-)11 }{2}   \\\\\\x_{1} =-8\\x_{2} =3\\\\9. x^{2} +2x+1=4x^{2} -4x+1\\3x^{2} -6x=0\\\\\\\\  \\\\x_{1} =0\\x_{2} =2\\\\3x(x-2)=0$