Answer :

[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.