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4q^2-18q+18=0
a = 4; b = -18; c = +18;
Δ = b2-4ac
Δ = -182-4·4·18
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*4}=\frac{12}{8} =1+1/2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*4}=\frac{24}{8} =3 $
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