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(3x-1)(4x-5)=0
We multiply parentheses ..
(+12x^2-15x-4x+5)=0
We get rid of parentheses
12x^2-15x-4x+5=0
We add all the numbers together, and all the variables
12x^2-19x+5=0
a = 12; b = -19; c = +5;
Δ = b2-4ac
Δ = -192-4·12·5
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*12}=\frac{8}{24} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*12}=\frac{30}{24} =1+1/4 $
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