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3x^2-4+2x=7x+2
We move all terms to the left:
3x^2-4+2x-(7x+2)=0
We get rid of parentheses
3x^2+2x-7x-2-4=0
We add all the numbers together, and all the variables
3x^2-5x-6=0
a = 3; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·3·(-6)
Δ = 97
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{97}}{2*3}=\frac{5-\sqrt{97}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{97}}{2*3}=\frac{5+\sqrt{97}}{6} $
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