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2x+x(x+10)=100
We move all terms to the left:
2x+x(x+10)-(100)=0
We multiply parentheses
x^2+2x+10x-100=0
We add all the numbers together, and all the variables
x^2+12x-100=0
a = 1; b = 12; c = -100;
Δ = b2-4ac
Δ = 122-4·1·(-100)
Δ = 544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{34}}{2*1}=\frac{-12-4\sqrt{34}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{34}}{2*1}=\frac{-12+4\sqrt{34}}{2} $
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