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6x-10=x^2-3
We move all terms to the left:
6x-10-(x^2-3)=0
We get rid of parentheses
-x^2+6x+3-10=0
We add all the numbers together, and all the variables
-1x^2+6x-7=0
a = -1; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·(-1)·(-7)
Δ = 8
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{2}}{2*-1}=\frac{-6-2\sqrt{2}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{2}}{2*-1}=\frac{-6+2\sqrt{2}}{-2} $
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