Algebra - Quadratic Equations

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Section 2.6 : Quadratic Equations - Part II

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12. Use the quadratic formula to solve the following quadratic equation.

\[169 - 20t + 4{t^2} = 0\]

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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.

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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 : \[\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{5}{2} - 6i\,\,{\mbox{and }}t = \frac{5}{2} + 6i}}\] .