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4x^2+9x+3=2x
We move all terms to the left:
4x^2+9x+3-(2x)=0
We add all the numbers together, and all the variables
4x^2+7x+3=0
a = 4; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·4·3
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*4}=\frac{-8}{8} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*4}=\frac{-6}{8} =-3/4 $
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