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x^2+11x=420
We move all terms to the left:
x^2+11x-(420)=0
a = 1; b = 11; c = -420;
Δ = b2-4ac
Δ = 112-4·1·(-420)
Δ = 1801
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{1801}}{2*1}=\frac{-11-\sqrt{1801}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1801}}{2*1}=\frac{-11+\sqrt{1801}}{2} $
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