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-3x-7+6x^2=7
We move all terms to the left:
-3x-7+6x^2-(7)=0
We add all the numbers together, and all the variables
6x^2-3x-14=0
a = 6; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·6·(-14)
Δ = 345
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{345}}{2*6}=\frac{3-\sqrt{345}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{345}}{2*6}=\frac{3+\sqrt{345}}{12} $
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