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2q^2-11q+12=18
We move all terms to the left:
2q^2-11q+12-(18)=0
We add all the numbers together, and all the variables
2q^2-11q-6=0
a = 2; b = -11; c = -6;
Δ = b2-4ac
Δ = -112-4·2·(-6)
Δ = 169
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*2}=\frac{-2}{4} =-1/2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*2}=\frac{24}{4} =6 $
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