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(7x-3)(3x+1)=0
We multiply parentheses ..
(+21x^2+7x-9x-3)=0
We get rid of parentheses
21x^2+7x-9x-3=0
We add all the numbers together, and all the variables
21x^2-2x-3=0
a = 21; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·21·(-3)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16}{2*21}=\frac{-14}{42} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16}{2*21}=\frac{18}{42} =3/7 $
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