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(4q-1)(3q+5)=0
We multiply parentheses ..
(+12q^2+20q-3q-5)=0
We get rid of parentheses
12q^2+20q-3q-5=0
We add all the numbers together, and all the variables
12q^2+17q-5=0
a = 12; b = 17; c = -5;
Δ = b2-4ac
Δ = 172-4·12·(-5)
Δ = 529
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{529}=23$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*12}=\frac{-40}{24} =-1+2/3 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*12}=\frac{6}{24} =1/4 $
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