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18x^2-36x+16=0
a = 18; b = -36; c = +16;
Δ = b2-4ac
Δ = -362-4·18·16
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12}{2*18}=\frac{24}{36} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12}{2*18}=\frac{48}{36} =1+1/3 $
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