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(3x+3)(7x-3)=99
We move all terms to the left:
(3x+3)(7x-3)-(99)=0
We multiply parentheses ..
(+21x^2-9x+21x-9)-99=0
We get rid of parentheses
21x^2-9x+21x-9-99=0
We add all the numbers together, and all the variables
21x^2+12x-108=0
a = 21; b = 12; c = -108;
Δ = b2-4ac
Δ = 122-4·21·(-108)
Δ = 9216
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{9216}=96$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-96}{2*21}=\frac{-108}{42} =-2+4/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+96}{2*21}=\frac{84}{42} =2 $
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