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x^2+12x+32=64
We move all terms to the left:
x^2+12x+32-(64)=0
We add all the numbers together, and all the variables
x^2+12x-32=0
a = 1; b = 12; c = -32;
Δ = b2-4ac
Δ = 122-4·1·(-32)
Δ = 272
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{17}}{2*1}=\frac{-12-4\sqrt{17}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{17}}{2*1}=\frac{-12+4\sqrt{17}}{2} $
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