2x^2+6x-179=0

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Solution for 2x^2+6x-179=0 equation:



2x^2+6x-179=0
a = 2; b = 6; c = -179;
Δ = b2-4ac
Δ = 62-4·2·(-179)
Δ = 1468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1468}=\sqrt{4*367}=\sqrt{4}*\sqrt{367}=2\sqrt{367}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{367}}{2*2}=\frac{-6-2\sqrt{367}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{367}}{2*2}=\frac{-6+2\sqrt{367}}{4} $

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