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x2+20x+91=0
We add all the numbers together, and all the variables
x^2+20x+91=0
a = 1; b = 20; c = +91;
Δ = b2-4ac
Δ = 202-4·1·91
Δ = 36
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6}{2*1}=\frac{-26}{2} =-13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6}{2*1}=\frac{-14}{2} =-7 $
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