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-2q-3=-2(2q+1)(3q)
We move all terms to the left:
-2q-3-(-2(2q+1)(3q))=0
We calculate terms in parentheses: -(-2(2q+1)3q), so:We get rid of parentheses
-2(2q+1)3q
We multiply parentheses
-12q^2-6q
Back to the equation:
-(-12q^2-6q)
12q^2+6q-2q-3=0
We add all the numbers together, and all the variables
12q^2+4q-3=0
a = 12; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·12·(-3)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*12}=\frac{-4-4\sqrt{10}}{24} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*12}=\frac{-4+4\sqrt{10}}{24} $
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