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5x+10x^2=20
We move all terms to the left:
5x+10x^2-(20)=0
a = 10; b = 5; c = -20;
Δ = b2-4ac
Δ = 52-4·10·(-20)
Δ = 825
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{33}}{2*10}=\frac{-5-5\sqrt{33}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{33}}{2*10}=\frac{-5+5\sqrt{33}}{20} $
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