Answer
$t=4.82,-5.82$
Work Step by Step
$t^2+t-28$
or, $t^2+t=28$
Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=1$
Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$
Thus, $c=\dfrac{b^2}{4a}=\dfrac{(1)^2}{4}=\dfrac{1}{4}$
To complete the square, add $\dfrac{1}{4}$ on both sides.
$t^2+t+\dfrac{1}{4}=28+\dfrac{1}{4}$
$\implies (t+\dfrac{1}{2})^2=\dfrac{113}{4}$
$\implies (t+\dfrac{1}{2})=5.32$
and
$\implies (t+\dfrac{1}{2})=-5.32$
or, $t=4.82,-5.82$
