-3y+y^2-18=0

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Solution for -3y+y^2-18=0 equation: 



-3y+y^2-18=0

a = 1; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*1}=\frac{-6}{2}
=-3
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*1}=\frac{12}{2}
=6
$

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