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5x^2-42x-400=0
a = 5; b = -42; c = -400;
Δ = b2-4ac
Δ = -422-4·5·(-400)
Δ = 9764
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9764}=\sqrt{4*2441}=\sqrt{4}*\sqrt{2441}=2\sqrt{2441}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{2441}}{2*5}=\frac{42-2\sqrt{2441}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{2441}}{2*5}=\frac{42+2\sqrt{2441}}{10} $
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