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x+(x+1)(x+2)=30
We move all terms to the left:
x+(x+1)(x+2)-(30)=0
We multiply parentheses ..
(+x^2+2x+x+2)+x-30=0
We get rid of parentheses
x^2+2x+x+x+2-30=0
We add all the numbers together, and all the variables
x^2+4x-28=0
a = 1; b = 4; c = -28;
Δ = b2-4ac
Δ = 42-4·1·(-28)
Δ = 128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{2}}{2*1}=\frac{-4-8\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{2}}{2*1}=\frac{-4+8\sqrt{2}}{2} $
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