X^2+2.4x-19.36=0

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Solution for X^2+2.4x-19.36=0 equation:



X^2+2.4X-19.36=0
a = 1; b = 2.4; c = -19.36;
Δ = b2-4ac
Δ = 2.42-4·1·(-19.36)
Δ = 83.2
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{-(2.4)-\sqrt{83.2}}{2*1}=\frac{-2.4-\sqrt{83.2}}{2} $
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.4)+\sqrt{83.2}}{2*1}=\frac{-2.4+\sqrt{83.2}}{2} $

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