4x2+16x+6=-0

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Solution for 4x2+16x+6=-0 equation:



4x^2+16x+6=-0
a = 4; b = 16; c = +6;
Δ = b2-4ac
Δ = 162-4·4·6
Δ = 160
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:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{10}}{2*4}=\frac{-16-4\sqrt{10}}{8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{10}}{2*4}=\frac{-16+4\sqrt{10}}{8} $

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