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3x(4x-2)=18
We move all terms to the left:
3x(4x-2)-(18)=0
We multiply parentheses
12x^2-6x-18=0
a = 12; b = -6; c = -18;
Δ = b2-4ac
Δ = -62-4·12·(-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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-30}{2*12}=\frac{-24}{24} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+30}{2*12}=\frac{36}{24} =1+1/2 $
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