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0=x^2-40x-560
We move all terms to the left:
0-(x^2-40x-560)=0
We add all the numbers together, and all the variables
-(x^2-40x-560)=0
We get rid of parentheses
-x^2+40x+560=0
We add all the numbers together, and all the variables
-1x^2+40x+560=0
a = -1; b = 40; c = +560;
Δ = b2-4ac
Δ = 402-4·(-1)·560
Δ = 3840
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3840}=\sqrt{256*15}=\sqrt{256}*\sqrt{15}=16\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{15}}{2*-1}=\frac{-40-16\sqrt{15}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{15}}{2*-1}=\frac{-40+16\sqrt{15}}{-2} $
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