Find the roots of the quadratic equation:?

Question

Find the roots of the quadratic equation:

w^2-3w-28=0
w=.
(If there is more than one root, separate them with commas.)

When I do these, I get massive sqrt and I think that it is excessive:
w=(-(-3)+((-3)^2-4*1*(-28))Ý2)/(2*1),
w=(-(-3)-((-3)^2-4*1*(-28))Ý2)/(2*1)

How is this supposed to be written?

Answer 1

If you work out the square (of -3) and the multiplications and divisions to find the value that you need to take the square root of, you will find that the number is 121. And you should be able to calculate the square root of 121.

It would be helpful to you if you developed your ability to factor trinomials like this, so that you could solve problems like this by factoring. I used to use the quadratic formula all the time, and in real-world problems it's often the only practical way to get an answer. But for school problems, many, many times you can do it much faster by factoring.

Answer 2

ax^2+bx-c=0

a,b, and c are all coefficients...and to use this method nicknamed "Mrs. Perron's Marvelous Method", a must equal 1. If not then you must use quadratic formula or completing the square. Links on how to use them are posted below.

therefore you must find factors of c (in this case c=28) . these factors when added must equal b.

therefore (-7) and (+4)

(x-7)(x+4)=0 therefore x= +7 or -4

Answer 3

Try and factor before using quadratic formula:

(w-7)(w+4)=0

w= +7 and -4