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2x^2+20x=100
We move all terms to the left:
2x^2+20x-(100)=0
a = 2; b = 20; c = -100;
Δ = b2-4ac
Δ = 202-4·2·(-100)
Δ = 1200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{3}}{2*2}=\frac{-20-20\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{3}}{2*2}=\frac{-20+20\sqrt{3}}{4} $
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