3x+x^2=240+2x

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Solution for 3x+x^2=240+2x equation:



3x+x^2=240+2x
We move all terms to the left:
3x+x^2-(240+2x)=0
We add all the numbers together, and all the variables
x^2+3x-(2x+240)=0
We get rid of parentheses
x^2+3x-2x-240=0
We add all the numbers together, and all the variables
x^2+x-240=0
a = 1; b = 1; c = -240;
Δ = b2-4ac
Δ = 12-4·1·(-240)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*1}=\frac{-32}{2} =-16 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*1}=\frac{30}{2} =15 $

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