-9y+y^2+8=0

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Solution for -9y+y^2+8=0 equation: 



-9y+y^2+8=0

a = 1; b = -9; c = +8;
Δ = b2-4ac
Δ = -92-4·1·8
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-7}{2*1}=\frac{2}{2}
=1
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+7}{2*1}=\frac{16}{2}
=8
$

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