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x+5x^2+x=1x^2+x-x-6x-1
We move all terms to the left:
x+5x^2+x-(1x^2+x-x-6x-1)=0
We add all the numbers together, and all the variables
5x^2+2x-(1x^2+x-x-6x-1)=0
We get rid of parentheses
5x^2-1x^2+2x-x+x+6x+1=0
We add all the numbers together, and all the variables
4x^2+8x+1=0
a = 4; b = 8; c = +1;
Δ = b2-4ac
Δ = 82-4·4·1
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*4}=\frac{-8-4\sqrt{3}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*4}=\frac{-8+4\sqrt{3}}{8} $
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