Question

I need you to answer with a, b, c, d

Answer 1

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

`\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}`

For the given function:

`f(x)=2x^2-10x-3`
`x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}`
`x=\frac{10\pm\sqrt[]{100+24}}{4}`
`\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}`
`\begin{gathered} x_1=(10)/(4)+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=(5)/(2)+\frac{\sqrt[]{31}}{2} \end{gathered}`
`\begin{gathered} x_2=(10)/(4)-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=(5)/(2)-\frac{\sqrt[]{31}}{2} \end{gathered}`

Then, the zeros of the given quadratic function are:

`\begin{gathered} x=(5)/(2)+\frac{\sqrt[]{31}}{2} \\ \\ x_{}=(5)/(2)-\frac{\sqrt[]{31}}{2} \end{gathered}`

Answer: Third option