0.16x^2+x-1.5=0

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Solution for 0.16x^2+x-1.5=0 equation:



0.16x^2+x-1.5=0
a = 0.16; b = 1; c = -1.5;
Δ = b2-4ac
Δ = 12-4·0.16·(-1.5)
Δ = 1.96
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{-(1)-\sqrt{1.96}}{2*0.16}=\frac{-1-\sqrt{1.96}}{0.32} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.96}}{2*0.16}=\frac{-1+\sqrt{1.96}}{0.32} $

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