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x2+16x=420
We move all terms to the left:
x2+16x-(420)=0
We add all the numbers together, and all the variables
x^2+16x-420=0
a = 1; b = 16; c = -420;
Δ = b2-4ac
Δ = 162-4·1·(-420)
Δ = 1936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-44}{2*1}=\frac{-60}{2} =-30 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+44}{2*1}=\frac{28}{2} =14 $
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