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12x^2+10x+2=0
a = 12; b = 10; c = +2;
Δ = b2-4ac
Δ = 102-4·12·2
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*12}=\frac{-12}{24} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*12}=\frac{-8}{24} =-1/3 $
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