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

$$a = 1\hspace{0.25in}b = 3\hspace{0.25in}c = - 130$$ Show Step 3

Plugging these into the quadratic formula gives,

$$z = \frac{{ - 3 \pm \sqrt {{{\left( 3 \right)}^2} - 4\left( 1 \right)\left( { - 130} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 3 \pm \sqrt {529} }}{2} = \frac{{ - 3 \pm 23}}{2}$$

In this case because there are no roots or complex numbers we can go further to reduce the solutions to “nicer” values.

$$z = \frac{{ - 3 - 23}}{2} = - 13,\,\,\,\,\,z = \frac{{ - 3 + 23}}{2} = 10$$

The two solutions to this equation are then : $$\require{bbox} \bbox[2pt,border:1px solid black]{{z = - 13\,\,{\mbox{and }}z = 10}}$$ .

As a final comment we can also note that because the solutions were integers we could have also gotten the answer by factoring! The quadratic (once written in standard form) factors as,

$$\left( {z + 13} \right)\left( {z - 10} \right) = 0$$

and this clearly would arrive at the same solutions.

The point of all this is to note that if more than one technique can be used it won’t matter which we use. Regardless of the solution technique used you will arrive at the same solutions.