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x^2-14x+7=10
We move all terms to the left:
x^2-14x+7-(10)=0
We add all the numbers together, and all the variables
x^2-14x-3=0
a = 1; b = -14; c = -3;
Δ = b2-4ac
Δ = -142-4·1·(-3)
Δ = 208
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{13}}{2*1}=\frac{14-4\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{13}}{2*1}=\frac{14+4\sqrt{13}}{2} $
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