(2x^2)-3x-5=0

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Solution for (2x^2)-3x-5=0 equation:



(2x^2)-3x-5=0
a = 2; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*2}=\frac{-4}{4} =-1 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*2}=\frac{10}{4} =2+1/2 $

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