Answer
$t=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
-2t(t+2)=-3
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-2t(t)-2t(2)=-3
\\\\
-2t^2-4t=-3
.\end{array}
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-2t^2-4t+3=0
\\\\
-1(-2t^2-4t+3)=-1(0)
\\\\
2t^2+4t-3=0
.\end{array}
The quadratic equation above has $a=
2
, b=
4
, c=
-3
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
t=\dfrac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)}
\\\\
t=\dfrac{-4\pm\sqrt{16+24}}{4}
\\\\
t=\dfrac{-4\pm\sqrt{40}}{4}
.\end{array}
Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to
\begin{array}{l}\require{cancel}
t=\dfrac{-4\pm\sqrt{4\cdot10}}{4}
\\\\
t=\dfrac{-4\pm\sqrt{(2)^2\cdot10}}{4}
\\\\
t=\dfrac{-4\pm2\sqrt{10}}{4}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
t=\dfrac{\cancel2(-2)\pm\cancel2(1)\sqrt{10}}{\cancel2(2)}
\\\\
t=\dfrac{-2\pm\sqrt{10}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
t=\dfrac{-2-\sqrt{10}}{2}
\\\\\text{OR}\\\\
t=\dfrac{-2+\sqrt{10}}{2}
.\end{array}
Hence, $
t=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2}
\right\}
.$