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(7x-4)(x-3)=0
We multiply parentheses ..
(+7x^2-21x-4x+12)=0
We get rid of parentheses
7x^2-21x-4x+12=0
We add all the numbers together, and all the variables
7x^2-25x+12=0
a = 7; b = -25; c = +12;
Δ = b2-4ac
Δ = -252-4·7·12
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-17}{2*7}=\frac{8}{14} =4/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+17}{2*7}=\frac{42}{14} =3 $
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