Question

hi guys this is the first time i have tryed this, can someone plz give me a step by step of working out x^2-4x+13=0

Answer 1

okay, what you need to do is find 2 numbers that multiplied are the last number (13), and added the second coefficient (-4)

Answer 2

if you cannot do that, then you need to add or subtract so that x²+2ax+a² is fullfilled

Answer 3

im pretty sure this wrong tbh u need to use -b+- sqrt b^2-4ac over 2

Answer 4

this leads to the formula maths nerd ish just posted

Answer 5

can u just do the formula bit plz
i know the answer just cant get to it

Answer 6

the prof of the formula, or how to use it?

Answer 7

how to use the formula

Answer 8

ok, simple
for a quadratic equation in the form ax²+bx+c= 0 , the solutions x1-2 are given by the formula: \[-b/(2a)\pm \sqrt{b²-4ac}\]

Answer 9

the sqare root should also be divided by 2a...

Answer 10

ik
i have said this already
you are taking 10 mins to type 3 lines ineed the in at some point plz

Answer 11

ok 4/2+-sqrt(16-4*13)/2

Answer 12

k,,,,,,

Answer 13

wow there is no point in this if you are going to take this long

Answer 14

Awfully ungrateful. Mik_wind's trying their best.

Answer 15

i don't know what you want me 2 do, if you already know the answers, and they are given by just typing that in the calculator

Answer 16

for the last 20 mins all i have been told is the simplist stuf to ughhhhhhhhhhhhhh

Answer 17

you should specify more exactly what you want

Answer 18

\[x^2-4x+13=0\]\[Δ=b^2-4ac\]\[Δ=(-4)^2-4\times1\times13=-36<0\]
This equation does not have any solutions in the set of real numbers.