3x(1+x)=13+x

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Solution for 3x(1+x)=13+x equation:



3x(1+x)=13+x
We move all terms to the left:
3x(1+x)-(13+x)=0
We add all the numbers together, and all the variables
3x(x+1)-(x+13)=0
We multiply parentheses
3x^2+3x-(x+13)=0
We get rid of parentheses
3x^2+3x-x-13=0
We add all the numbers together, and all the variables
3x^2+2x-13=0
a = 3; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·3·(-13)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{10}}{2*3}=\frac{-2-4\sqrt{10}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{10}}{2*3}=\frac{-2+4\sqrt{10}}{6} $

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