Question

What are the factors of the square root of 329

Answer 1

Answer: $x=-\dfrac{3}{2}, 7$

Step-by-step explanation:

$2x^2-10x-26=x-5$

$2x^2-10x-26-x+5=0$  Move all terms to one side

$2x^2-11x-21=0$ Simplify $2{x}^{2}-10x-26-x+5$

$2x^2+3x-14x-21=0$ Split the second term

$x(2x+3)-7(2x+3)=0$  Factor out common terms in the first two terms, then in the last two terms.

$(2x+3)(x-7)=0$ Factor out the common term $2x+3$

$x=-\dfrac{3}{2}, 7$

Answer 2

Answer:

x = -3/4 and 7/2 will be your final answer, but check below for a correction on your work

Step-by-step explanation:

***Your work is incorrect, it should be

2x² - 11x - 21 = 0

you need to subtract an 'x' from both sides, and add 5 to both sides...you added an x to the left side and subtracted a 5 from the left side instead***

I'll solve this so in case you will need help on it too.

Now you'll have

x = (11 ± √[(-11²) - 4(2)(-21)])/([4(2)]

x = (11 ±  √289)/8

x = (11 ± 17)/8

x = (11 + 17)/8 = 28/8 = 7/2

or

x = (11 - 17)/8 = -6/8 = -3/4