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2x-1x-3-3(x-4)2=-x2+17x-45
We move all terms to the left:
2x-1x-3-3(x-4)2-(-x2+17x-45)=0
We add all the numbers together, and all the variables
-(-1x^2+17x-45)+2x-1x-3(x-4)2-3=0
We add all the numbers together, and all the variables
-(-1x^2+17x-45)+x-3(x-4)2-3=0
We multiply parentheses
-(-1x^2+17x-45)+x-6x+24-3=0
We get rid of parentheses
1x^2-17x+x-6x+45+24-3=0
We add all the numbers together, and all the variables
x^2-22x+66=0
a = 1; b = -22; c = +66;
Δ = b2-4ac
Δ = -222-4·1·66
Δ = 220
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{55}}{2*1}=\frac{22-2\sqrt{55}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{55}}{2*1}=\frac{22+2\sqrt{55}}{2} $
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