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x2-19x+3=0
We add all the numbers together, and all the variables
x^2-19x+3=0
a = 1; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·1·3
Δ = 349
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{-(-19)-\sqrt{349}}{2*1}=\frac{19-\sqrt{349}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{349}}{2*1}=\frac{19+\sqrt{349}}{2} $
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