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x2+16x=-41
We move all terms to the left:
x2+16x-(-41)=0
We add all the numbers together, and all the variables
x^2+16x+41=0
a = 1; b = 16; c = +41;
Δ = b2-4ac
Δ = 162-4·1·41
Δ = 92
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{23}}{2*1}=\frac{-16-2\sqrt{23}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{23}}{2*1}=\frac{-16+2\sqrt{23}}{2} $
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