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(7-6x)(1+3x)=3+21x
We move all terms to the left:
(7-6x)(1+3x)-(3+21x)=0
We add all the numbers together, and all the variables
(-6x+7)(3x+1)-(21x+3)=0
We get rid of parentheses
(-6x+7)(3x+1)-21x-3=0
We multiply parentheses ..
(-18x^2-6x+21x+7)-21x-3=0
We get rid of parentheses
-18x^2-6x+21x-21x+7-3=0
We add all the numbers together, and all the variables
-18x^2-6x+4=0
a = -18; b = -6; c = +4;
Δ = b2-4ac
Δ = -62-4·(-18)·4
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-18}=\frac{-12}{-36} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-18}=\frac{24}{-36} =-2/3 $
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