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12x^2+48x-1=0
a = 12; b = 48; c = -1;
Δ = b2-4ac
Δ = 482-4·12·(-1)
Δ = 2352
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-28\sqrt{3}}{2*12}=\frac{-48-28\sqrt{3}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+28\sqrt{3}}{2*12}=\frac{-48+28\sqrt{3}}{24} $
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