2h2+9h+8=0

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Solution for 2h2+9h+8=0 equation: 



2h^2+9h+8=0

a = 2; b = 9; c = +8;
Δ = b2-4ac
Δ = 92-4·2·8
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{17}}{2*2}=\frac{-9-\sqrt{17}}{4}
$


$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{17}}{2*2}=\frac{-9+\sqrt{17}}{4}
$

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