Step-by-step explanation:
The easiest way to solve this (in my opinion) is to use the Pythagorean theorem [tex]\frac{-b+-\sqrt{b^2-4(a)(c)} }{2a}[/tex]
In this case, 5 is a, with the highest power, 13 is b, with the second highest power, and -6 is c, with a 0 power.
Or, looking at this equation [tex]ax^2+bx+c[/tex] we can gain that knowledge as well.
So let's plug these values in and solve!
[tex]\frac{-b+-\sqrt{b^2-4(a)(c)} }{2a} = \frac{-13+-\sqrt{13^2-4(5)(-6)} }{2(5)}\\=\frac{-13+-\sqrt{169+120} }{10}=\frac{-13+-\sqrt{289} }{10}=\frac{-13+-17 }{10}[/tex]
Now we simplify this into two different answers, one when we add, and one when we subtract:
[tex]add:\frac{-13+17 }{10}=\frac{4}{10}=\frac{2}{5}\\subtract:\frac{-13-17 }{10}=\frac{-30 }{10}=-3[/tex]
So, assuming the question asked to solve for x (which I'm assuming it did), our answers are x = -3 and [tex]\frac{2}{5}[/tex]
Answer:
x = -3
x = [tex]\frac{2}{5}[/tex]
