0.5x^2+4x+4.5=0

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Solution for 0.5x^2+4x+4.5=0 equation: 



0.5x^2+4x+4.5=0

a = 0.5; b = 4; c = +4.5;
Δ = b2-4ac
Δ = 42-4·0.5·4.5
Δ = 7
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{-(4)-\sqrt{7}}{2*0.5}=\frac{-4-\sqrt{7}}{1}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{7}}{2*0.5}=\frac{-4+\sqrt{7}}{1}
$

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