(7x^2+6x)+(x^2+6x=)

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Solution for (7x^2+6x)+(x^2+6x=) equation: 



(7x^2+6x)+(x^2+6x=)

We move all terms to the left:

(7x^2+6x)+(x^2+6x-())=0

We get rid of parentheses

7x^2+6x+(x^2+6x-())=0



We calculate terms in parentheses: +(x^2+6x-()), so:

x^2+6x-()

We add all the numbers together, and all the variables

x^2+6x

Back to the equation:

+(x^2+6x)



We get rid of parentheses

7x^2+x^2+6x+6x=0

We add all the numbers together, and all the variables

8x^2+12x=0

a = 8; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·8·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*8}=\frac{-24}{16}
=-1+1/2
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*8}=\frac{0}{16}
=0
$

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