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b2+14b+45=0
We add all the numbers together, and all the variables
b^2+14b+45=0
a = 1; b = 14; c = +45;
Δ = b2-4ac
Δ = 142-4·1·45
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*1}=\frac{-18}{2} =-9 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*1}=\frac{-10}{2} =-5 $
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