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10x^2+20x-10=0
a = 10; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·10·(-10)
Δ = 800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*10}=\frac{-20-20\sqrt{2}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*10}=\frac{-20+20\sqrt{2}}{20} $
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