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12x^2+16x+2=0
a = 12; b = 16; c = +2;
Δ = b2-4ac
Δ = 162-4·12·2
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*12}=\frac{-16-4\sqrt{10}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{10}}{2*12}=\frac{-16+4\sqrt{10}}{24} $
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