5x^2-4+x=0

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Solution for 5x^2-4+x=0 equation: 



5x^2-4+x=0

a = 5; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·5·(-4)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*5}=\frac{-10}{10}
=-1
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*5}=\frac{8}{10}
=4/5
$

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