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5q^2+26q=0
a = 5; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·5·0
Δ = 676
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{676}=26$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*5}=\frac{-52}{10} =-5+1/5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*5}=\frac{0}{10} =0 $
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