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10x-26=x2-4
We move all terms to the left:
10x-26-(x2-4)=0
We add all the numbers together, and all the variables
-(+x^2-4)+10x-26=0
We get rid of parentheses
-x^2+10x+4-26=0
We add all the numbers together, and all the variables
-1x^2+10x-22=0
a = -1; b = 10; c = -22;
Δ = b2-4ac
Δ = 102-4·(-1)·(-22)
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{3}}{2*-1}=\frac{-10-2\sqrt{3}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{3}}{2*-1}=\frac{-10+2\sqrt{3}}{-2} $
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