4x^2+2x=100

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Solution for 4x^2+2x=100 equation:



4x^2+2x=100
We move all terms to the left:
4x^2+2x-(100)=0
a = 4; b = 2; c = -100;
Δ = b2-4ac
Δ = 22-4·4·(-100)
Δ = 1604
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:
$\sqrt{\Delta}=\sqrt{1604}=\sqrt{4*401}=\sqrt{4}*\sqrt{401}=2\sqrt{401}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{401}}{2*4}=\frac{-2-2\sqrt{401}}{8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{401}}{2*4}=\frac{-2+2\sqrt{401}}{8} $

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