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0=3x^2+3x-4
We move all terms to the left:
0-(3x^2+3x-4)=0
We add all the numbers together, and all the variables
-(3x^2+3x-4)=0
We get rid of parentheses
-3x^2-3x+4=0
a = -3; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-3)·4
Δ = 57
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{57}}{2*-3}=\frac{3-\sqrt{57}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*-3}=\frac{3+\sqrt{57}}{-6} $
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