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a2+7a+3=3
We move all terms to the left:
a2+7a+3-(3)=0
We add all the numbers together, and all the variables
a^2+7a=0
a = 1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·1·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*1}=\frac{-14}{2} =-7 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*1}=\frac{0}{2} =0 $
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