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x^2-18(x-4)=0
We multiply parentheses
x^2-18x+72=0
a = 1; b = -18; c = +72;
Δ = b2-4ac
Δ = -182-4·1·72
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*1}=\frac{12}{2} =6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*1}=\frac{24}{2} =12 $
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