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16x^2+48x+28=0
a = 16; b = 48; c = +28;
Δ = b2-4ac
Δ = 482-4·16·28
Δ = 512
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{2}}{2*16}=\frac{-48-16\sqrt{2}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{2}}{2*16}=\frac{-48+16\sqrt{2}}{32} $
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