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scZoUnD [109]
3 years ago
14

Does anyone know how to do this?

Mathematics
2 answers:
Vesnalui [34]3 years ago
6 0

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.

                                    <u>6x² = -19x - 15</u>

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

Add  15  to each side:    <u>6x² + 19x + 15 = 0</u>

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 ]

<em>x = - (1 and 2/3)</em>         or   <em>x = -(1 and 1/2)</em> 

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.


Vsevolod [243]3 years ago
6 0
6x^2=-19x-15\\&#10;6x^2+19x+15=0\\&#10;6x^2+10x+9x+15=0\\&#10;2x(3x+5)+3(3x+5)=0\\&#10;(2x+3)(3x+5)=0\\&#10;2x+3=0 \vee 3x+5=0\\&#10;2x=-3 \vee 3x=-5\\&#10;x=-\dfrac{3}{2} \vee x=-\dfrac{5}{3}&#10;
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