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6x^2+7x=1
We move all terms to the left:
6x^2+7x-(1)=0
a = 6; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·6·(-1)
Δ = 73
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{73}}{2*6}=\frac{-7-\sqrt{73}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{73}}{2*6}=\frac{-7+\sqrt{73}}{12} $
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