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x2+10x+100=225
We move all terms to the left:
x2+10x+100-(225)=0
We add all the numbers together, and all the variables
x^2+10x-125=0
a = 1; b = 10; c = -125;
Δ = b2-4ac
Δ = 102-4·1·(-125)
Δ = 600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{6}}{2*1}=\frac{-10-10\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{6}}{2*1}=\frac{-10+10\sqrt{6}}{2} $
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