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3x^2+3x-4=-2
We move all terms to the left:
3x^2+3x-4-(-2)=0
We add all the numbers together, and all the variables
3x^2+3x-2=0
a = 3; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·3·(-2)
Δ = 33
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{-(3)-\sqrt{33}}{2*3}=\frac{-3-\sqrt{33}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*3}=\frac{-3+\sqrt{33}}{6} $
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