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-3(x)=x^2-1
We move all terms to the left:
-3(x)-(x^2-1)=0
We get rid of parentheses
-x^2-3x+1=0
We add all the numbers together, and all the variables
-1x^2-3x+1=0
a = -1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·(-1)·1
Δ = 13
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{13}}{2*-1}=\frac{3-\sqrt{13}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*-1}=\frac{3+\sqrt{13}}{-2} $
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