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x^2-6x-10=13
We move all terms to the left:
x^2-6x-10-(13)=0
We add all the numbers together, and all the variables
x^2-6x-23=0
a = 1; b = -6; c = -23;
Δ = b2-4ac
Δ = -62-4·1·(-23)
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{-(-6)-8\sqrt{2}}{2*1}=\frac{6-8\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+8\sqrt{2}}{2*1}=\frac{6+8\sqrt{2}}{2} $
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