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c2+10c+16=0
We add all the numbers together, and all the variables
c^2+10c+16=0
a = 1; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*1}=\frac{-16}{2} =-8 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*1}=\frac{-4}{2} =-2 $
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