(2x^2)+9x-5=0

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Solution for (2x^2)+9x-5=0 equation: 



(2x^2)+9x-5=0

a = 2; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·2·(-5)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*2}=\frac{-20}{4}
=-5
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*2}=\frac{2}{4}
=1/2
$

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