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3x^2+3=7x
We move all terms to the left:
3x^2+3-(7x)=0
a = 3; b = -7; c = +3;
Δ = b2-4ac
Δ = -72-4·3·3
Δ = 13
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{-(-7)-\sqrt{13}}{2*3}=\frac{7-\sqrt{13}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{13}}{2*3}=\frac{7+\sqrt{13}}{6} $
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