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-1300-220x+x^2=0
a = 1; b = -220; c = -1300;
Δ = b2-4ac
Δ = -2202-4·1·(-1300)
Δ = 53600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{53600}=\sqrt{400*134}=\sqrt{400}*\sqrt{134}=20\sqrt{134}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-220)-20\sqrt{134}}{2*1}=\frac{220-20\sqrt{134}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-220)+20\sqrt{134}}{2*1}=\frac{220+20\sqrt{134}}{2} $
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