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(3a-7)(a+14)=(2a+1)
We move all terms to the left:
(3a-7)(a+14)-((2a+1))=0
We multiply parentheses ..
(+3a^2+42a-7a-98)-((2a+1))=0
We calculate terms in parentheses: -((2a+1)), so:We get rid of parentheses
(2a+1)
We get rid of parentheses
2a+1
Back to the equation:
-(2a+1)
3a^2+42a-7a-2a-98-1=0
We add all the numbers together, and all the variables
3a^2+33a-99=0
a = 3; b = 33; c = -99;
Δ = b2-4ac
Δ = 332-4·3·(-99)
Δ = 2277
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2277}=\sqrt{9*253}=\sqrt{9}*\sqrt{253}=3\sqrt{253}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{253}}{2*3}=\frac{-33-3\sqrt{253}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{253}}{2*3}=\frac{-33+3\sqrt{253}}{6} $
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