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x^2+16x=32
We move all terms to the left:
x^2+16x-(32)=0
a = 1; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·1·(-32)
Δ = 384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{6}}{2*1}=\frac{-16-8\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{6}}{2*1}=\frac{-16+8\sqrt{6}}{2} $
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