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]
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.