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(Q)=3Q^2+5Q-2
We move all terms to the left:
(Q)-(3Q^2+5Q-2)=0
We get rid of parentheses
-3Q^2+Q-5Q+2=0
We add all the numbers together, and all the variables
-3Q^2-4Q+2=0
a = -3; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·(-3)·2
Δ = 40
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}$$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*-3}=\frac{4-2\sqrt{10}}{-6} $$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-3}=\frac{4+2\sqrt{10}}{-6} $
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