Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.