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12x^2+25x+2=0
a = 12; b = 25; c = +2;
Δ = b2-4ac
Δ = 252-4·12·2
Δ = 529
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*12}=\frac{-48}{24} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*12}=\frac{-2}{24} =-1/12 $
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