x^2+0.5x-14=0

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Solution for x^2+0.5x-14=0 equation:



x^2+0.5x-14=0
a = 1; b = 0.5; c = -14;
Δ = b2-4ac
Δ = 0.52-4·1·(-14)
Δ = 56.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{56.25}}{2*1}=\frac{-0.5-\sqrt{56.25}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{56.25}}{2*1}=\frac{-0.5+\sqrt{56.25}}{2} $

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