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1 year ago

All you need is in the photo I DON'T WANT STEP BY STEP ANSWER FAST please fdsd

Mathematics

1 answer:

Brut [27]1 year ago

6 0

We have the following:

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=5 \\ b=0 \\ c=-80 \end{gathered}$

replacing:

$\begin{gathered} x=\frac{-0\pm\sqrt[]{0^2-4\cdot5\cdot-80}}{2\cdot5}=\frac{\pm\sqrt[]{1600}}{10}=\frac{\pm40}{10}=\pm4 \\ x_1=4 \\ x_2=-4 \end{gathered}$

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