2*x^2+5x-48=0

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Solution for 2*x^2+5x-48=0 equation: 



2x^2+5x-48=0

a = 2; b = 5; c = -48;
Δ = b2-4ac
Δ = 52-4·2·(-48)
Δ = 409
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{-(5)-\sqrt{409}}{2*2}=\frac{-5-\sqrt{409}}{4}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{409}}{2*2}=\frac{-5+\sqrt{409}}{4}
$

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