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5x^2-40x+50=0
a = 5; b = -40; c = +50;
Δ = b2-4ac
Δ = -402-4·5·50
Δ = 600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{6}}{2*5}=\frac{40-10\sqrt{6}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{6}}{2*5}=\frac{40+10\sqrt{6}}{10} $
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