Решите уравнение;подробное решение 1)121-x^2=0 2)18x+6x^2=0 3)11x^2-12=0 4)x^2-9x+10=0...

1) $121- x^{2} =0 \\ - x^{2} =-121\\ x^{2} =121\\x= \sqrt{121} \\x_{1} = 11\\ x_{2} = -11$

2) $18x+6 x^{2} =0\\6x(3+x)=0\\x=0\\3+x=0\\x=-3$

3) $11 x^{2} -12=0\\11 x^{2} =12\\ x^{2} = \frac{12}{11} \\ x_{1} = \sqrt{ \frac{12}{11} } = \sqrt{1 \frac{1}{11} } \\ x_{2} =- \sqrt{ \frac{12}{11} }=- \sqrt{1 \frac{1}{11} }$

4) $x^{2} -9x+10=0\\ D= b^{2} -4ac\\D= 9^{2} -4*10=41\\ x_{1}= \frac{-b+ \sqrt{D} }{2a}= \frac{9+ \sqrt{41} }{2} \\ x_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{9- \sqrt{41} }{2}$

5) $3 x^{2} -7x+5=0\\D=7^{2} -4*5*3=49-60=-11$
Корней нет

6) $-2 x^{2} +8x+5=0\\2 x^{2} -8x-5=0\\D=8^{2} -4*(-5)*2=64-40=24\\ x_{1} = \frac{8+ \sqrt{24} }{4} = \frac{8+2 \sqrt{6} }{4} = \frac{4+ \sqrt{6} }{2} \\ x_{2} = \frac{8- \sqrt{24} }{4}= \frac{8-2 \sqrt{6} }{4}= \frac{4- \sqrt{6} }{2}$