2y2-18y+40=0

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Solution for 2y2-18y+40=0 equation: 



2y^2-18y+40=0

a = 2; b = -18; c = +40;
Δ = b2-4ac
Δ = -182-4·2·40
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*2}=\frac{16}{4}
=4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*2}=\frac{20}{4}
=5
$

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