-3(t)=-2t^2+1

Solution for -3(t)=-2t^2+1 equation:

-3(t)=-2t^2+1

We move all terms to the left:

-3(t)-(-2t^2+1)=0

We get rid of parentheses

2t^2-3t-1=0

a = 2; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·2·(-1)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{17}}{2*2}=\frac{3-\sqrt{17}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{17}}{2*2}=\frac{3+\sqrt{17}}{4}
$