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5x^2+40x-400=0
a = 5; b = 40; c = -400;
Δ = b2-4ac
Δ = 402-4·5·(-400)
Δ = 9600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9600}=\sqrt{1600*6}=\sqrt{1600}*\sqrt{6}=40\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{6}}{2*5}=\frac{-40-40\sqrt{6}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{6}}{2*5}=\frac{-40+40\sqrt{6}}{10} $
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