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12x^2+7x=13
We move all terms to the left:
12x^2+7x-(13)=0
a = 12; b = 7; c = -13;
Δ = b2-4ac
Δ = 72-4·12·(-13)
Δ = 673
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{673}}{2*12}=\frac{-7-\sqrt{673}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{673}}{2*12}=\frac{-7+\sqrt{673}}{24} $
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