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2y^2-18y+8=0
a = 2; b = -18; c = +8;
Δ = b2-4ac
Δ = -182-4·2·8
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{65}}{2*2}=\frac{18-2\sqrt{65}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{65}}{2*2}=\frac{18+2\sqrt{65}}{4} $
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