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x+4x^2=2+5x
We move all terms to the left:
x+4x^2-(2+5x)=0
We add all the numbers together, and all the variables
4x^2+x-(5x+2)=0
We get rid of parentheses
4x^2+x-5x-2=0
We add all the numbers together, and all the variables
4x^2-4x-2=0
a = 4; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·4·(-2)
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{-(-4)-4\sqrt{3}}{2*4}=\frac{4-4\sqrt{3}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*4}=\frac{4+4\sqrt{3}}{8} $
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