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2x+42x^2=
We move all terms to the left:
2x+42x^2-()=0
We add all the numbers together, and all the variables
42x^2+2x=0
a = 42; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·42·0
Δ = 4
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*42}=\frac{-4}{84} =-1/21 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*42}=\frac{0}{84} =0 $
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