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2x^2-36x+18=0
a = 2; b = -36; c = +18;
Δ = b2-4ac
Δ = -362-4·2·18
Δ = 1152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-24\sqrt{2}}{2*2}=\frac{36-24\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+24\sqrt{2}}{2*2}=\frac{36+24\sqrt{2}}{4} $
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