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X2+13X+42=0
We add all the numbers together, and all the variables
X^2+13X+42=0
a = 1; b = 13; c = +42;
Δ = b2-4ac
Δ = 132-4·1·42
Δ = 1
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{1}=1$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*1}=\frac{-14}{2} =-7 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*1}=\frac{-12}{2} =-6 $
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