Question

1) 5x^2-36x+7=0 2) 7x^2+20x-3=0 3) 5x^2-26x+5=0 4) 8x^2+15x-2=0 5) x^2-x-6=0 Решите пожалуйста

Answer 1

1)$5 x^{2} -36x+7=0 \\ k=36/2=18 \\ D= 18^{2} - 5*7=324-35=289 \\ x_{1} = \frac{-18- \sqrt{289} }{5} = -7 \\ x_{2} = \frac{-18+ \sqrt{289} }{5} = -0.2$
2)$7x^{2} +20x-3=0 \\ k=20/2=10 \\ D= 10^{2} -7*(-3)=100+21=121 \\ x_{1} = \frac{-10- \sqrt{121} }{7} = -3 \\ x_{2} = \frac{-10+ \sqrt{121} }{7} = \frac{1}{7}$
3)$5x^{2} -26x+5=0 \\ k=26/2=13 \\ D= 13^{2} -5*5=169-25=144 \\ x_{1} = \frac{-13- \sqrt{144} }{5} = - 5 \\ x_{2} = \frac{-13+ \sqrt{144} }{5} = - 0.2$
4)$8x^{2} =15x-2=0 \\ D= 15^{2} - 4*8*(-2)=225-(-64)=289 \\ x_{1} = \frac{-15- \sqrt{289} }{16} = -2 \\ x_{2} = \frac{-15+ \sqrt{289} }{16} =0.125$
5)$x^{2} -x-6=0 \\ D =1^{2} -4*1*(-6)=1-(-24)=25 \\ x_{1} = \frac{1+ \sqrt{25} }{2} =3 \\ x_{2} = \frac{1- \sqrt{25} }{2} = - 2$