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4x^2+3x+2x=6
We move all terms to the left:
4x^2+3x+2x-(6)=0
We add all the numbers together, and all the variables
4x^2+5x-6=0
a = 4; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·4·(-6)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*4}=\frac{6}{8} =3/4 $
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