-0.1x^2+10=0

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Solution for -0.1x^2+10=0 equation: 



-0.1x^2+10=0

a = -0.1; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-0.1)·10
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*-0.1}=\frac{-2}{-0.2}
=+10
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*-0.1}=\frac{2}{-0.2}
=-10
$

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