We are given the equation
[tex]5t^2-6t=4[/tex]
To solve this question using the quadratic formula
We will re-write the equation as follow:
[tex]5t^2-6t-4=0[/tex]
Next, we will compare with the general quadratic formula
[tex]ax^2+bx+c=0[/tex]
In our case
[tex]a=5,b=-6,c=-4[/tex]
To apply the quadratic formula
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
What we will do next will be to substitute the values of a,b, and c
[tex]t=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\times5\times-4}}{2\times5}[/tex]
[tex]t=\frac{6\pm\sqrt[]{36+80}}{10}[/tex]
[tex]t=\frac{6\pm\sqrt[]{116}}{10}[/tex]
[tex]t=\frac{6\pm2\sqrt[]{29}}{2\times5}[/tex]
Simplifying further
[tex]\begin{gathered} t=\frac{6+2\sqrt[]{29}}{2\times5},t=\frac{6-2\sqrt[]{29}}{2\times5} \\ \\ t=\frac{3+\sqrt[]{29}}{5},t=\frac{3-\sqrt[]{29}}{5} \end{gathered}[/tex]
If we are to get the values in decimals, the value of t will be
[tex]t=1.67703,\: t=-0.47703[/tex]
