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x^2+14x-10=8x-6
We move all terms to the left:
x^2+14x-10-(8x-6)=0
We get rid of parentheses
x^2+14x-8x+6-10=0
We add all the numbers together, and all the variables
x^2+6x-4=0
a = 1; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·1·(-4)
Δ = 52
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{13}}{2*1}=\frac{-6-2\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{13}}{2*1}=\frac{-6+2\sqrt{13}}{2} $
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