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q2+7q+9=0
We add all the numbers together, and all the variables
q^2+7q+9=0
a = 1; b = 7; c = +9;
Δ = b2-4ac
Δ = 72-4·1·9
Δ = 13
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{13}}{2*1}=\frac{-7-\sqrt{13}}{2} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{13}}{2*1}=\frac{-7+\sqrt{13}}{2} $
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