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10x^2=50
We move all terms to the left:
10x^2-(50)=0
a = 10; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·10·(-50)
Δ = 2000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*10}=\frac{0-20\sqrt{5}}{20} =-\frac{20\sqrt{5}}{20} =-\sqrt{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*10}=\frac{0+20\sqrt{5}}{20} =\frac{20\sqrt{5}}{20} =\sqrt{5} $
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