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X2-19X+48=0
We add all the numbers together, and all the variables
X^2-19X+48=0
a = 1; b = -19; c = +48;
Δ = b2-4ac
Δ = -192-4·1·48
Δ = 169
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{169}=13$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*1}=\frac{6}{2} =3 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*1}=\frac{32}{2} =16 $
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