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6x^2=1-4x

We move all terms to the left:

6x^2-(1-4x)=0

We add all the numbers together, and all the variables

6x^2-(-4x+1)=0

We get rid of parentheses

6x^2+4x-1=0

a = 6; b = 4; c = -1;

Δ = b^{2}-4ac

Δ = 4^{2}-4·6·(-1)

Δ = 40

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{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*6}=\frac{-4-2\sqrt{10}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*6}=\frac{-4+2\sqrt{10}}{12} $

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