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3x^2+x=4
We move all terms to the left:
3x^2+x-(4)=0
a = 3; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·3·(-4)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*3}=\frac{-8}{6} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*3}=\frac{6}{6} =1 $
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