5x squared plus 13x-6=0

Question

asked Oct 25, 2020 178k views

2 Answers

Kevin Van Mierlo · Answer 1 · 2020-10-26T09:38:06+0000

Explanation:

The easiest way to solve this (in my opinion) is to use the Pythagorean theorem
(-b+-√(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!

$(-b+-√(b^2-4(a)(c)) )/(2a) = (-13+-√(13^2-4(5)(-6)) )/(2(5))\\=(-13+-√(169+120) )/(10)=(-13+-√(289) )/(10)=(-13+-17 )/(10)$

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

$add:(-13+17 )/(10)=(4)/(10)=(2)/(5)\\subtract:(-13-17 )/(10)=(-30 )/(10)=-3$

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

Answer:

x = -3

x =
(2)/(5)