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x2+22x+120.5=0
We add all the numbers together, and all the variables
x^2+22x+120.5=0
a = 1; b = 22; c = +120.5;
Δ = b2-4ac
Δ = 222-4·1·120.5
Δ = 2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-\sqrt{2}}{2*1}=\frac{-22-\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+\sqrt{2}}{2*1}=\frac{-22+\sqrt{2}}{2} $
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