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(x)(x+1)=42
We move all terms to the left:
(x)(x+1)-(42)=0
We multiply parentheses
x^2+x-42=0
a = 1; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·1·(-42)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*1}=\frac{-14}{2} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*1}=\frac{12}{2} =6 $
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