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(2a-4)+2(a-5)=-3a(a+1)
We move all terms to the left:
(2a-4)+2(a-5)-(-3a(a+1))=0
We multiply parentheses
(2a-4)+2a-(-3a(a+1))-10=0
We get rid of parentheses
2a+2a-(-3a(a+1))-4-10=0
We calculate terms in parentheses: -(-3a(a+1)), so:We add all the numbers together, and all the variables
-3a(a+1)
We multiply parentheses
-3a^2-3a
Back to the equation:
-(-3a^2-3a)
-(-3a^2-3a)+4a-14=0
We get rid of parentheses
3a^2+3a+4a-14=0
We add all the numbers together, and all the variables
3a^2+7a-14=0
a = 3; b = 7; c = -14;
Δ = b2-4ac
Δ = 72-4·3·(-14)
Δ = 217
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{217}}{2*3}=\frac{-7-\sqrt{217}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{217}}{2*3}=\frac{-7+\sqrt{217}}{6} $
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