Question

Use the quadratic formula to find the exact solutions of the following equation

Answer 1

We are given the equation

`5t^2-6t=4`

To solve this question using the quadratic formula

We will re-write the equation as follow:

`5t^2-6t-4=0`

Next, we will compare with the general quadratic formula

`ax^2+bx+c=0`

In our case

`a=5,b=-6,c=-4`

To apply the quadratic formula

`t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

What we will do next will be to substitute the values of a,b, and c

`t=\frac{-(-6)\pm\sqrt[]{(-6)^2-4*5*-4}}{2*5}`

`t=\frac{6\pm\sqrt[]{36+80}}{10}`

`t=\frac{6\pm\sqrt[]{116}}{10}`

`t=\frac{6\pm2\sqrt[]{29}}{2*5}`

Simplifying further

`\begin{gathered} t=\frac{6+2\sqrt[]{29}}{2*5},t=\frac{6-2\sqrt[]{29}}{2*5} \\ \\ t=\frac{3+\sqrt[]{29}}{5},t=\frac{3-\sqrt[]{29}}{5} \end{gathered}`

If we are to get the values in decimals, the value of t will be

`t=1.67703,\: t=-0.47703`