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x(x+1)+(x+2)+(x+3)=130
We move all terms to the left:
x(x+1)+(x+2)+(x+3)-(130)=0
We multiply parentheses
x^2+x+(x+2)+(x+3)-130=0
We get rid of parentheses
x^2+x+x+x+2+3-130=0
We add all the numbers together, and all the variables
x^2+3x-125=0
a = 1; b = 3; c = -125;
Δ = b2-4ac
Δ = 32-4·1·(-125)
Δ = 509
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{-(3)-\sqrt{509}}{2*1}=\frac{-3-\sqrt{509}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{509}}{2*1}=\frac{-3+\sqrt{509}}{2} $
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