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(2x+3)(6x-1)=0
We multiply parentheses ..
(+12x^2-2x+18x-3)=0
We get rid of parentheses
12x^2-2x+18x-3=0
We add all the numbers together, and all the variables
12x^2+16x-3=0
a = 12; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·12·(-3)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*12}=\frac{-36}{24} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*12}=\frac{4}{24} =1/6 $
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