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2x(3+x)=9
We move all terms to the left:
2x(3+x)-(9)=0
We add all the numbers together, and all the variables
2x(x+3)-9=0
We multiply parentheses
2x^2+6x-9=0
a = 2; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·2·(-9)
Δ = 108
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{3}}{2*2}=\frac{-6-6\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{3}}{2*2}=\frac{-6+6\sqrt{3}}{4} $
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