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(5x-4)(x-3)=0
We multiply parentheses ..
(+5x^2-15x-4x+12)=0
We get rid of parentheses
5x^2-15x-4x+12=0
We add all the numbers together, and all the variables
5x^2-19x+12=0
a = 5; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·5·12
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*5}=\frac{8}{10} =4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*5}=\frac{30}{10} =3 $
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