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x^2+6x+9=40
We move all terms to the left:
x^2+6x+9-(40)=0
We add all the numbers together, and all the variables
x^2+6x-31=0
a = 1; b = 6; c = -31;
Δ = b2-4ac
Δ = 62-4·1·(-31)
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{-(6)-4\sqrt{10}}{2*1}=\frac{-6-4\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{10}}{2*1}=\frac{-6+4\sqrt{10}}{2} $
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