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4q^2+9q-5=5q-5
We move all terms to the left:
4q^2+9q-5-(5q-5)=0
We get rid of parentheses
4q^2+9q-5q+5-5=0
We add all the numbers together, and all the variables
4q^2+4q=0
a = 4; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·4·0
Δ = 16
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{16}=4$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*4}=\frac{-8}{8} =-1 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*4}=\frac{0}{8} =0 $
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