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10x^2+2x=368
We move all terms to the left:
10x^2+2x-(368)=0
a = 10; b = 2; c = -368;
Δ = b2-4ac
Δ = 22-4·10·(-368)
Δ = 14724
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14724}=\sqrt{36*409}=\sqrt{36}*\sqrt{409}=6\sqrt{409}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{409}}{2*10}=\frac{-2-6\sqrt{409}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{409}}{2*10}=\frac{-2+6\sqrt{409}}{20} $
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