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### Section 2-6 : Quadratic Equations - Part II

$169 - 20t + 4{t^2} = 0$

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First, we need to get the quadratic equation in standard form. This is,

$4{t^2} - 20t + 169 = 0$

Note that in this case we just rearranged the terms to have decreasing exponents.

Show Step 2

Now we need to identify the values for the quadratic formula.

$a = 4\hspace{0.25in}b = - 20\hspace{0.25in}c = 169$ Show Step 3

Plugging these into the quadratic formula gives,

$t = \frac{{ - \left( { - 20} \right) \pm \sqrt {{{\left( { - 20} \right)}^2} - 4\left( 4 \right)\left( {169} \right)} }}{{2\left( 4 \right)}} = \frac{{20 \pm \sqrt { - 2304} }}{8} = \frac{{20 \pm 48i}}{8} = \frac{5}{2} \pm 6i$

The two solutions to this equation are then : $\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{5}{2} - 6i\,\,{\mbox{and }}t = \frac{5}{2} + 6i}}$ .