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c2+16c+60=0.
We add all the numbers together, and all the variables
c^2+16c+60=0
a = 1; b = 16; c = +60;
Δ = b2-4ac
Δ = 162-4·1·60
Δ = 16
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{16}=4$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*1}=\frac{-20}{2} =-10 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*1}=\frac{-12}{2} =-6 $
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