Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Does anyone know how to do this?

Does Anyone Know How To Do This class=

Sagot :

[tex]6x^2=-19x-15\\ 6x^2+19x+15=0\\ 6x^2+10x+9x+15=0\\ 2x(3x+5)+3(3x+5)=0\\ (2x+3)(3x+5)=0\\ 2x+3=0 \vee 3x+5=0\\ 2x=-3 \vee 3x=-5\\ x=-\dfrac{3}{2} \vee x=-\dfrac{5}{3} [/tex]
AL2006

Well, first of all, you haven't told us what you want to do with it, and
there are no instructions in the picture.

But I do see a quadratic equation in the picture, which can probably be
solved to find the values of 'x' that make the equation a true statement.

                                    6x² = -19x - 15

Add  19x  to each side:   6x² + 19x = -15

Add  15  to each side:    6x² + 19x + 15 = 0

With the equation now in standard form, you can either try and factor the
left side, or else do it the easy way and apply the quadratic formula.

x = (1/12) [ -19 plus or minus √(19² - 360) ]

x = (1/12) [ -19 plus or minus √1 ]

x = (1/12) [ -20 ]  or  x = (1/12) [ -18 ]

x = - (1 and 2/3)         or   x = -(1 and 1/2) 

That's what you can do with the equation in the picture. 
I don't know what you can do with the two crossed arrows.


Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.