Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
[tex]15x^2-1=2x\\\\15x^2-2x-1=0\\\\a=15;\ b=-2;\ c=-1\\\\\Delta=b^2-4ac\to\Delta=(-2)^2-4\cdot15\cdot(-1)=4+60=64\\\\x_1=\frac{-b-\sqrt\Delta}{2}a;\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{64}=8\\\\x_1=\frac{2-8}{2\cdot15}=\frac{-6}{30}=-\frac{1}{5};\ x_2=\frac{2+8}{2\cdot15}=\frac{10}{30}=\frac{1}{3}[/tex]
[tex]15x^2 - 1 = 2x \\\\15x^2-2x - 1 = 0\\ \\a=15, \ \ b=-2 , \ \ c=-1 \\ \\\Delta =b^2-4ac = (-2)^2 -4\cdot 15 \cdot (-1) = 4+60=64 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-(-2)-\sqrt{64}}{2\cdot 15 }=\frac{ 2-8}{30}=\frac{-6}{30}=- \frac{1}{5}\\ \\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-(-2)+\sqrt{64}}{2\cdot 15 }=\frac{ 2+8}{30}=\frac{10}{30}= \frac{1}{3}
[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.