4r^2-5=59

Question

4r^2-5=59

anonymous · Answer

question is ? : 
$$4r^2-5=59$$

anonymous · Answer

right?

anonymous · Answer

Yeah. I've tried to do the squared symbol and I can't

anonymous · Answer

oh, a'ight. that isn't of concern. u can use this formula:
$$r=(-b \pm \sqrt{b^2-4ac})/2a$$

anonymous · Answer

are u familiar with this?

anonymous · Answer

Lol. No!

anonymous · Answer

okay, i'll explain it :3 !

anonymous · Answer

Okay (:

anonymous · Answer

in a typical quadratic equation:
$$ax^2 +bx +c =0$$

anonymous · Answer

u can use that formula to get the roots of the equation. now considering the equation at hand:
$$4r^2 -5=59$$

anonymous · Answer

where a=4, b=0 (since there is no 'r' term we consider that to be:
$$4r^2+0r-5=59$$

texaschic101 · Answer

add 5 to both sides....then divide by 4.....then take square root of that answer

anonymous · Answer

which eventually gives us this:
$$4r^2+0r-64=0$$

anonymous · Answer

a=4 b=0 & c=-64

anonymous · Answer

use that in the formula & u will get the answer this way too:
$$r=\frac{(-0 \pm \sqrt{0-4*4*(-64)})}{2*4}$$

anonymous · Answer

@LezBeHonest_69

texaschic101 · Answer

4x^2 - 5 = 59 --- add 5 to both sides
4x^2 = 59 + 5
4x^2 = 64 -- divide both sides by 4
x^2 = 64/4
x^2 = 16
x = sq rt 16
x = (+ - 4)

check..
4x^2 - 5 = 59
4(4^2) - 4 = 59
4(16) - 5 = 59
64 - 5 = 59
59 = 59 (correct)

It would also be correct with x being -4, because a -4^2 is also 16