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200-20x-5x^2=0
a = -5; b = -20; c = +200;
Δ = b2-4ac
Δ = -202-4·(-5)·200
Δ = 4400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4400}=\sqrt{400*11}=\sqrt{400}*\sqrt{11}=20\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{11}}{2*-5}=\frac{20-20\sqrt{11}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{11}}{2*-5}=\frac{20+20\sqrt{11}}{-10} $
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