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12x^2+42x+18=0
a = 12; b = 42; c = +18;
Δ = b2-4ac
Δ = 422-4·12·18
Δ = 900
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{900}=30$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-30}{2*12}=\frac{-72}{24} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+30}{2*12}=\frac{-12}{24} =-1/2 $
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