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2x+16x^2=100
We move all terms to the left:
2x+16x^2-(100)=0
a = 16; b = 2; c = -100;
Δ = b2-4ac
Δ = 22-4·16·(-100)
Δ = 6404
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6404}=\sqrt{4*1601}=\sqrt{4}*\sqrt{1601}=2\sqrt{1601}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1601}}{2*16}=\frac{-2-2\sqrt{1601}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1601}}{2*16}=\frac{-2+2\sqrt{1601}}{32} $
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