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x^2+16x=11
We move all terms to the left:
x^2+16x-(11)=0
a = 1; b = 16; c = -11;
Δ = b2-4ac
Δ = 162-4·1·(-11)
Δ = 300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-10\sqrt{3}}{2*1}=\frac{-16-10\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+10\sqrt{3}}{2*1}=\frac{-16+10\sqrt{3}}{2} $
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