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q2+49q=0
We add all the numbers together, and all the variables
q^2+49q=0
a = 1; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·1·0
Δ = 2401
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{2401}=49$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*1}=\frac{-98}{2} =-49 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*1}=\frac{0}{2} =0 $
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