Question

5x squared plus 13x-6=0​

Answer 1

Use the quadratic equation: you get values of x=2/5 and x=-3

Answer 2

Step-by-step explanation:

The easiest way to solve this (in my opinion) is to use the Pythagorean theorem $\frac{-b+-\sqrt{b^2-4(a)(c)} }{2a}$

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 $ax^2+bx+c$ we can gain that knowledge as well.

So let's plug these values in and solve!

$\frac{-b+-\sqrt{b^2-4(a)(c)} }{2a} = \frac{-13+-\sqrt{13^2-4(5)(-6)} }{2(5)}\\=\frac{-13+-\sqrt{169+120} }{10}=\frac{-13+-\sqrt{289} }{10}=\frac{-13+-17 }{10}$

Now we simplify this into two different answers, one when we add, and one when we subtract:

$add:\frac{-13+17 }{10}=\frac{4}{10}=\frac{2}{5}\\subtract:\frac{-13-17 }{10}=\frac{-30 }{10}=-3$

So, assuming the question asked to solve for x (which I'm assuming it did), our answers are x = -3 and $\frac{2}{5}$

Answer:

x = -3

x = $\frac{2}{5}$