Z(t)=2t2+7t-4

Solution for Z(t)=2t2+7t-4 equation:

(Z)=2Z^2+7Z-4

We move all terms to the left:

(Z)-(2Z^2+7Z-4)=0

We get rid of parentheses

-2Z^2+Z-7Z+4=0

We add all the numbers together, and all the variables

-2Z^2-6Z+4=0

a = -2; b = -6; c = +4;
Δ = b2-4ac
Δ = -62-4·(-2)·4
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{17}}{2*-2}=\frac{6-2\sqrt{17}}{-4}
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{17}}{2*-2}=\frac{6+2\sqrt{17}}{-4}
$