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w2+14w+48=0
We add all the numbers together, and all the variables
w^2+14w+48=0
a = 1; b = 14; c = +48;
Δ = b2-4ac
Δ = 142-4·1·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*1}=\frac{-16}{2} =-8 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*1}=\frac{-12}{2} =-6 $
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