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12x^2+40x+8=0
a = 12; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·12·8
Δ = 1216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{19}}{2*12}=\frac{-40-8\sqrt{19}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{19}}{2*12}=\frac{-40+8\sqrt{19}}{24} $
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