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12x^2+16x-4=0
a = 12; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·12·(-4)
Δ = 448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{7}}{2*12}=\frac{-16-8\sqrt{7}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{7}}{2*12}=\frac{-16+8\sqrt{7}}{24} $
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